UNIVERSITY  OF  CALIFORNIA 
LOS  ANGELES 


GIFT  OF 

Col.  Glen  F. 


ORDNANCE:  DERARTI^IEINT  DOCUMENT   NO.  2O3S-A 


THEORY  AND  DE5IGN 

OF 

RECOIL   5Y5TEM5 

AND 

GUN  CARRIAGES 


(CHAPTER     3ZHI) 


Prepared  in  the 
oT  -the    ChieF  aF   Ordnance 


JANUARY 


CONFIDENTIAL 


ENGINEER  REF3  RODUCTION  F=l_ANT 
WASHINGTON    BARR  ACKS.O.C. 


UP 
LL 


ORDNANCE  DEPARTMENT 
Document  No.2035-A 
Office  of  the  Chief  of  Ordnance 


WAP  DEPARTMENT, 

Washington,  February  1922. 

The  following  publication,  entitled  "Theory  and 
Design  of  Recoil  Systems  and  Gun  Carriages",  is 
published  for  the  information  and  guidance  of  all 
students  of  the  Ordnance  training  schools,  military 
academies,  and  other  similar  educational  organizations, 

The  contents  are  strictly  of  a  Confidential 
nature  and  should  not  be  reputlished  without  authority 
of  this  Department. 

C.  C.  WILLIAMS, 
Major  General, Chief  of  Ordnance,  U.S.A. 


523414 


HYDRO  PNEUMATIC  RECOIL  SYSTEMS. 


INTRODUCTION.       The  fundamental  advantages  of 

these  recoil  systems  are  smooth- 
ness of  operation  with  conse- 
quent durability  and  ability 
to  meet  with  requirements  of 

service  within  space  and  weight  limitations.   Light 
and  small  recuperator  forgings  are  made  possible 
by  the  use  of  "high  pressures,  both  in  the  re- 
cuperator cylinder  as  well  as  in  the  hydraulic. 
To  obtain  and  hold  these  pressures,  special  sys- 
tems of  packing  are  necessary  and  successful 
manufacture  depends  largely  upon  use  of  proper 
packing. 

RECOIL  SYSTEM  WITH  This  system  consists  usually 
CONSTANT  ORIFICE    of  two  cylinders;  the  hydraulic 
FIXED  LENGTH  OF     brake  cylinder  and  a  recuperator 
RECOIL.  cylinder  containing  a  floating 

piston  which  separates  the  oil 

and  air,  this  cylinder  communicating  with  the  hy- 
draulic cylinder  by  means  of  a  large  opening  at  one 
end  of  the  cylinder. 

The  floating  piston  has  a  long  buffer  rod  which 
passes  through  a  fixed  diaphragm  in  the  recuperator 
cylinder  and  enters  a  buffer  at  its  end,  serving  a 
double  purpose  since  it  gives  constant  throttling 
through  the  diaphragm  on  recoil  and  serves  as  a 
throttling  buffer  on  counter  recoil. 

During  recoil  the  constant  orifice  about  the 
buffer  rod  at  the  diaphragm  of  the  recuperator 
cylinder  produces  a  throttling  of  the  oil  and  a 
consequent  drop  of  pressure  due  to  the  constant 
orifice. 

3 


During  counter  recoil  the  oil  passes  through 
a  constant  orifice  produced  by  constant  throttling 
grooves  in  the  buffer  chamber.   Toward  the  end  of 
recoil  the  depth  of  these  grooves  vary,  causing  a 
varying  orifice.   The  throttling  is  so  designed  as 
to  cause  a  sufficient  drop  of  pressure  from  the 
air,  to  produce  a  pressure  against  the  hydraulic 
piston  just  sufficient  to  balance  the  total 
friction  and  thus  nearly  maintain  uniform  motion 
through  a  considerable  part  of  counter  recoil. 

At  the  end  of  c 'recoil  where  we  have  the  de- 
creasing variable  orifice,  the  pressure  is  rapidly 
reduced  to  zero;  thus  the  recoiling  mass  is  brought 
to  rest  practically  by  only  the  total  friction  of 
the  guides,  stuffing  box  and  piston  frictions. 

A  characteristic  of  %  constant  orifice  in  a 
recoil  system  is  to  give  a  peak  in  the  pressure 
curve  in  the  region  of  the  maximum  velocity  of  re- 
coil.  The  maximum  velocity  and  the  consequent 
maximum  drop  of  pressure  due  to  throttling  occurs, 
fortunately  at  the  beginning  of  recoil  where  the 
air  pressure  is  at  its  minimum.   As  the  gun  recoils, 
the  drop  of  pressure  decreases  due  to  the  decreased 
velocity  and  consequent  less  throttling  through 
the  orifice,  whereas  the  air  pressure  rises,  and 
since  at  all  times  the  hydraulic  pressure  is  the 
sum  of  the  air  and  drop  of  pressure  due  to  throttling, 
we  have  a  tendency  to  maintain  a  constant  hydraulic 
pressure  in  the  brake  cylinder.  This  effect  is 
especially  true  with  a  fairly  long  recoil.   If, 
however,  the  recoil  is  shortened,  the  peak  effect 
due  to  the  throttling  at  the  beginning  of  recoil 
is  much  greater  than  the  increase  of  air  pressure 
at  the  end  of  recoil. 

The  type  of  system  now  under  discussion  has  a 
single  length  of  recoil  for  all  elevations  which  is  made 
sufficiently  long  for  stability  at  zero  degrees, 
while  the  ratio  of  final  air  pressure  to  initial 


air  pressure  and  the  constant  orifice  are  so  de- 
signed as  to  give  a  uniform,  or  nearly  uniform 
pressure  curve  consistent  with  the  stability  slope. 
This  type  is  ohviously  restricted  to  a  limited  ele- 
vation. 

It  is  proposed  now  to  discuss  the  theory  de- 
sign both  for  recoil  and  counter  recoil  and  then 
to  approximate  design  formulae  for  the  use  in  de- 
sign,  Le  t 

.A  =  effective  area  of  the  hydraulic  or  brake 

piston 

Aa  =  area  of  the  air  side  of  the  floating  piston, 
A^  =  effective  area  of  the  oil  side  of  the 

floating  piston, 
at,  =  area  of  the  buffer  rod. 

0Ajj  =  area  of  the  buffer  chamber  or  buffer  head. 
Wo  =  area  of  constant  orifice  during  the  re- 
coil, 

W^  =  area  of  counter  recoil  constant  orifice. 
Wx  =  variable  orifice  in  buffer  chamber  during 

end  of  counter  recoil. 

p  =  intensity  of  pressure  against  hydraulic  pis- 
ton 

pt  =  intensity  of  ressure  in  recuperator 
cylinder  back  of  valve  or  at  entrance  to 
constant  orifice. 

Pt-=  drop  of  pressure  in  passing  spring  valve. 
P  »  drop  of  pressure  in  throttling  through 

constant  orifice. 
Rv  =  reaction  of  spiral  spring  a-fe  lift  "h"  on 

«0 

spring  valve. 

d -=  spring  constant  of  the  spiral  springs 
ho  =  initial  compression  of  spring 
as  =  area  of  hase  of  spring  valve 

c  =  circumference  of  spring  valve  «  rrd 
If  K  equals  the  total  resistance  to  recoil,  we  have 

during  the  powder  pressure  period,  the  total  powder 
force  being  Pt 


J  2  y  F—  R  P  K 

F-K=M_  -   hence  —  r-  At=Av   or  -r-At-  ~  At=Av 
r  dt2         Mr          Mr    Mr 

bat  for  the  same  time  intervals  when  the  recoil- 
ing mass  if  free,  assuming  the  powder  force  to  be 

unaffected  by  the  slightly  different  motion  of  the 
powder  charge  and  shell,  we  have 

F 

—  At=Av*  where  v*  as  the  free  velocity  of  recoil 

\i 

Mr  for  the  recoiling  mass. 

Hence  during  the  powder  period, 

f{ 
A  Vf  -  —  -  At=Av  and  the  corresponding  displacement 

of  recoil,  becomes,  A  x  =  v  A  t. 
During  the  subsequent  retardation, 

J  2  yr  V 

-  K=Mr  —  -   or  -  —  -  A  t  =  A  v,  when  K  =  pA+R-Wrsin# 
r  and  R  =  the  total 

guide  and 
stuffing  box  and  piston  friction. 

Considering  now  figure(l)  we  have  the  pres- 
sure at  the  base  of  the  spring  valve  equal  to  the 
pressure  against  the  hydraulic  piston,  neglecting 
the  slight  velocity  head  generated  at  the  base  of 
the  valve. 

From  the  law  of  continuity,  the  quantity  of 
oil  passing  through  the  valves,  becomes, 

Qv=AV-AbV-'  where  V  =  the  velocity  of  the  float- 

AV 
ing  piston  =  — 

Aa 

.Ab 
thus  Qv  =  AV(1  -  —  )     Now  if  we  neglect  the 

A 

dynamic  reaction  against 

the  valve  which  is  small,  we  have  Pjas=Rv=S(h+ho) 
but  P  for  the  four  valves,  becomes, 


p  =     -         =  --•••••        hence   the   valve   lift 
2800  (ch)*  as  is   obtained   from   the 

cubic  equation: 


h*(h+b)  *  —————   and  the  corresponding  drop 

of   pressure  by  substituting 


the  K  obtained  from  the  above  equation  in  the  formula 

P~Pi=Pi  =  ~  <h*ho> 
as 

In  general  this  refinement  may  be  omitted,  since 
the  drop  through  the  valves  is  at  a  maximum,  but  10* 
of  the  total  drop,  and  the  movement  of  the  spring  is 
at  a  maximum  not  greater  than  one  third  its  total 
deflection  to  solid  height,  thus  making  a  gross 

error  of  less  than  3*  if  we  consider  the  spring  re- 
action that  at  assembled  height.  Consequently  in 
the  calculation  of  the  pressure  and  velocity  curves 
we  will  assume  Pta=S  ho=p-pi=  a  constant. 

Now  the  quantity  of  oil  passing  through  the 
constant  orifice,  is  equal  to  that  passing  through 
the  valves  plus  the  quantity  of  oil  displaced  by  the 

annular  area  about  the  buffer  rod  in  the  buffer 
chamber,  that  is  Q0-(AV-AbV  '  )+  (Afc-a^V1  =  AV-abV 

ab       A' 
=  AV(1  -- 


A  check  on  this  valve  may  be  obtained  by  noting 
that  of  the  total  quantity  of  oil  displaced  by  the 

floating  piston  AaV'=AV  but  (Aa-ab)V'  =  AaV'  passes 

AV 
through  the  constant  orifice,  and  we  note  V  =  —  • 

Aa 

Aa~ab 
hence  as  above,  QQ  =  (-•  •-•  )AV 

Aa 

The  drop  of  pressure  due  to  throttling  through 
the  constant  orifice,  becomes, 

K*  d« 
po  =  n  n     =P  ~Pa   Therefore,  we  have,  p=PJ*Po+pl 


° 

To  obtain  p^  we  must  further  consider  the  equilibrium 
of  the  floating  piston,  neglecting  the  effect  of  its 
inertia  during  the  acceleration  and  retardation  as 


being  small.  The  forces  in  the  direction  of  motion 
acting  on  the  floating  piston,  are: 

(1)  The  air  pressure  on  the  air  side 
of  the  piston  resisting  the  motion. 

(2)  The  packing  friction  resisting 
motion. 

(3)  The  oil  pressure  on  the  oil  side 
of  the  floating  piston, 

(4)  A  resisting  pressure  on  the  annular 
area  of  the  buffer. 

(5)  The  pressure  on  the  base  of  the 
buffer. 

The  pressure  in  the  buffer  chamber  is  assumed 
the  same  as  the  pressure  in  the  hydraulic  or  recoil 
cylinder  since  by  computation  it  is  found  that  the 
drop  in  pressure  due  to  throttling  through  the  small 
set  of  holes  at  the  buffer  end  of  the  recuperator 
cylinder  is  exceedingly  small. 

Therefore,  considering  the  equilibrium  of  the 
floating  piston  and  neglecting  the  drop  of  pressure 
through  the  valves,  we  have 

^  -  paAa-Rf=0 


Aa  Rf,   ab 
hence  pa  =  ——  —  —  —  —  =(p  —  >  —  )-p  — 


The  pressure   in  the  hydraulic  or   recoil   cylinder 
becomes, 

K0(Aa-ab)«A«V«  Aa     Rf  ab 

p=Sh«  +   —  —  —  ^—  —  —  •*•   p  —  —  +  —  -  p  — 
"«  175   A|w0  Aa      Aa  Aa 

therefore 

ab  K*Aa2A2Va  Aa  Rf 

p(l  +  —  )  =Sh  +  -  +  p*  where  Pa=Pa"~+~" 

A^         175A«w«  A^  Aa 

The  resistance  to  recoil,  becomes,  since 


10 


ah   Aa 
—  )~ 

Aa  Aa 


ShoAaA 
"      ^ 


176A|w« 


or  in  terms  of  pa, 


where 

Rs  =  stuffing  box  friction  of  piston  rod. 

R_  =  hydraulic  piston  friction 
Rg  =  guide  friction 

As  will  be  shown  in  the  theory  of  packing  the 
stuffing  box  and  hydraulic  piston  frictions  de- 

pend upon  the  pressure  in  the  hydraulic  cylinder 
Further  a  recoil  buffer  action  is  obtained 

towards  the  end  of  the  recoil  by  making  WQ 

variable,  i.e.  substituting  wx  in  the  above 

equation  for  WQ. 

The  total  resistance  to  recoil  may  be 

simplified  into  the  following  equation: 

V* 
K=Cjpa+Cf  -7  +(ZR+Cg)-W4sin# 

Towards  the  end  of  recoil  HO  changes  to  wx 

i 


1*  A3 
A 


K 


R(A 


*  s 

The  above  equation  is  exactly  equivalent  to 
previous  recoil  equations  with  the  fundamental 
difference  that  now  we  have  a  constant  orifice  during 
the  greater  part  of  recoil  in  place  of  the  previous 
variable  orifice. 

If  the  term  Ca—  •  is  large  as  compared  with 
0   the  other  terms  in  the 


11 


equation  for  the  resistance  to  recoil,  K  obviously 
varies  considerably  with  the  velocity  of  recoil, 
having  a  large  peak  at  maximum  velocity.   To  reduce 
this  peak  in  the  resistance  curve,  we  must  have  a 
larger  orifice  with  less  throttling  and  a  conse- 
quent longer  recoil  together  with  a  higher  air  pres- 
sure than  would  be  used  on  other  types  of  recoil 
systems.   Hence,  we  may  completely  limit  the  ratio 
of  the  peak  to  the  average  resistance  to  recoil  by 
suitable  high  air  pressure,  a  large  constant  orifice 
and  a  long  recoil.   Thus  we  see  the  limitation  of 

this  system  for  high  elevation  and  consequqnt  short 
recoil. 

THEORY  OF  PACKING.       To  prevent  leakage  of  air 

or  oil  under  high  pres- 
sures in  the  cylinders 
the  packing. is  designed  to 
be  subjected  to  the  actual 

intensity  of  pressure  of  the  air  or  oil  as  the  case 
may  be.   The  packing,  however,  cannot  be  regarded  as 
subjected  exactly  to  a  hydraulic  pressure  since  the 
pressure  in  one  direction  differs  from  the  pressure 
in  a  direction  at  right  angles  by  the  amount  of  the 
previous  pressure  multiplied  by  Poisson  Ratio.  To 
compensate  for  the  deficiency  in  oil  or  air  pres- 
sure normal  to  the  cylinder  surface  or  piston  rod, 
packing  springs  are  introduced.   These  springs  are 
designed  for  a  slightly  excess  pressure  normal  to 
the  cylinder  surface  in  addition  to  compensating 
for  the  air  or  oil  hydrostatic  pressure.   The  total 
normal  force  multiplied  by  the  coefficient  of  friction 

of  the  packing  gives  the  total  friction.   Thus  we 

see  the  friction  may  be  considered  as  consisting 
of  two  components;  the  friction  component  due  to 
the  oil  or  air,  and  that  due  to  the  packing  springs. 
The  packing  frictions,  therefore,  are  a  linear 
function  of  the  pressure  and  may  be  represented  by 
an  equation  as  follows:   F  =ciP+ct 


12 


1 
ct  =  gross  length  of  outer  flaps  of  flange 

K  =  Poissons1  ratio  for  leather  =  0.73 

pa  =  intensity  of  air  pressure 

Pg  =  intensity  of  oil  side  of  floating  piston 

p0  =  intensity  of  pressure  in  leather  on  air 
side  caused  by  packing  springs. 

PQ  =  intensity  of  pressure  in  leather  on  oil 
side  caused  by  packing  springs. 

pq  =  intensity  of  pressure  in  grease  caused  by 
reaction  of  spiral  spring  on  slide  moving 
on  inner  rod  of  floating  piston. 

Rs  =  reaction  of  spiral  spring  on  grease 

slide  at  assembled  height. 
Rb  =  reaction  of  belleville  washers  on  annular 

area  of  packing  on  air  side  at  assembled 

height. 
Kb  =  reaction  of  belleville  washers  on  oil 

side  of  packing  at  assembled  height, 
f  =  coefficient  of  friction  of  leather  =  .05 
ft  =  coefficient  of  silver  =  .09 

Then  in  the  metric  system  of  units  if  all  lengths  or 
dimensions  are  measured  in  millimeters  and  pressures 
in  kilograms  per  sq.  cm.  and  spring  reaction  in 

kilograms, 

400  Rs  400  Rb 

PO  "      .  _  » 


400Rb«  n 

p1  =  ——   where  -(d*-d2 )  is  the  an- 
n(d*-da-)        4 

nular  of  the 

packing  ring 

in  sq.  mm.  and  the  total  pressure  in  the  packing, 
become, 

Grease  pressure  =  Pa+Pq  (  considered  a  purely 

hydrostatic  pres- 
sure) 


13 


I 
i 


14 


Air      side  packing  pressure  =  pa  +    po        Kg   per 

sq.    cm. 
Oil   side   packing   pressure   =  Pa  *   PQ 

If  we   multiply   the   packing   pressures  by  K  we   obtain 
the   pressure   normal    to   the   cylinder  surface. 
Prom   the  diagram  of   packing   contact  with  cylinder 
fig.    (2)   we   nay  assume 

=  c-2e  and  a  =c'-2  e 

£f 

since  the  contact  with  cylinder  is  some  fraction  of 
the  breadth  of  flap  of  the  flange. 

The  total  friction  of  the  floating  piston  pack- 
ing becomes,  in  kilograms, 

Ff  • 


if  jPq].01K  +  7td[  (bf+af  t 


*  Trd[(bf+afi)(p0+p0)+2aifipq]  .01K+  nd  [2(bf  +af  J 

^a^pa  .OIK 

Thus  the  packing  friction  reduces  to  Rf=CJ+C9pa 
(kilograms)  where  Ct-  irdt  (bf  +  af  x  )  (po+po=+2atf  tpq]  .OIK 

and  Cf  =  ndt2(bf+afi)+2  atf13  .OIK 

The  drop  of  pressure  due  to  the  packipg  friction 
of  the  floating  piston,  becomes,  in  Ibs.  per  sq. 

inch  2.205-Rf 

±    (Pa"Pa^=pf=  ""7  -       very  roughly   (See  previous 


exact  formula) 
Where  Aa  =  the  area  of  the  floating  piston  in  sq. 
inches  and  the  plus  sign  for  recoil  and  minus 
for  counter  recoil. 

The  above  formulae  are  not  strictly  exact  but 
give  however,  a  very  close  approximation,  indeed, 
of  the  friction  drop  due  to  the  floating  piston. 

In  the  design  of  the  packing  given  later, 
tables  are  shown  for  proper  width  of  packing  and  ex- 
cess pressures  due  to  springs  as  well  as  a  table  for 


15 


Belleville  given  in  the  appendix. 

STUFFING  BOX  FRICTION.   In  the  stuffing  box  packing 

the  leather  and  flaps  of  the 
packing  flanges,  come  in 
contact  with  the  peripheral 
surface  of  the  piston  rod. 

A  typical  stuffing  box  packing  is  shown  in  figure  (3). 
If,  now,  as  before 

po  =  intensity  of  pressure  caused  by  bellevilles 

or  packing  springs  in  Kg.  per  sq.  cm. 
p  =  pressure  in  hydraulic  cylinder  Kg.  per  sq. 

cm. 

dr  =  diam.  of  piston  rod  in  millimeters. 
do  =  outer  diameter  of  packing  ring  in  millimeters 
RJ.J  =  Belleville  reaction  at  assembled  load  on 
annular  area  of  packing  ring  measured  in 
kilograms 
Then  in  metric  units  as  before, 

400   Eb  tr 

p   =      (A*  _d2>     where  7   (d*-d*)is   the   annular   area  of 
7r\Q0~Qr  ^  4- 

the  packing  ring 

and  the  stuffing 
box  friction,  becomes, 

Rs  =  nd[  (bf+afi)(pQ+p)+a1f  tp]  .OIK  Kilograms 
=  irdCbfp+afp  +  a 


]  .OIK 

Or  as  before  Rs=Ci+C2p  where  Ct  =  *  dpo(bf  +af  ±)  .C1K 

and  C9  =  7td(bf+af+  at£t).01K 

HYDRAULIC  PISTON  FRICTION.   The  packing  in  the 

hydraulic  piston  is 
similar  to  that  in  the 
floating  piston.   We  have,  if 

d  =  the  diam.  of  the  hydraulic  cylinder  in  mm. 
d  =  the  inner  diameter  of  the  packing  ring, 
in  mm. 


16 


R^  =  the  reaction  of  the  belleville  washers  on 
the  packing,  in  Kg. 
400  Rb 

P°   n(da-d») 
or  in  the  form  R_  =  Ci  +  CJtp 

Ct  =  ndpoCbf  +  af^.OlK 

Cf  =  nd(bf+aft+  a^)  .OIK 

CALCULATED  RETARDED  VELOCITY       During  the  pow- 
CURVE  OF  A  TYPICAL  SYSTEM.         der  period,  we 

have 


or  in  terms  of  any  two  given  points, 
K(t  -t  ) 


v-v 


vfra  vfn  vm  vn 

mr 

During  the  retardation,  we  have  V  = 
.'.  V  -Vm  = 


The  corresponding  displacement  during  the  interval, 
becomes, 

Xmn=  A  X, 


and  Pa  =  Pa~*JT 
s 

Friction  Values: 

(1)     Floating  piston  friction 

K  =  0.73      Rs  =  950  Ibs.       RB  =  2150  Ibs. 

RjJ=  4080  Ib.   p-  =  190  Ibs.  per  sq.  in.  (calculated  ) 

p0  =  880  Ibs.  per  sq.in.          p,J  =  1690  Ibs.  per  sq.in, 


17 


Hence  from  the  previous  theory  on  packing,  the 
floating  piston  friction  may  "be  expressed  in  the 
following  formula:  Rf  =  .53  pa+420  Ibs. 

(2)  Hydraulic  piston  packing  friction 
Rb  =  2980    P0  =  2080  Ibs.  per  sq.in, 

Hence  R  =  .121  p  +280 

(3)  Stuffing  box  friction 

Rb=2810  Ibs.  p0=1970  Ibs.  per  sq.in. 
Hence  RS  =  .063  p  +  140 

(4)  Guide  friction  neglected 

From  (2)  and- (3 )  we  obtain  for  the 
total  friction  resistance  during  the 
recoil,  SR=.184  p+420 
Drop  of  pressure  through  check  valve: 

as  =  0.0779  sq.in.     Sho  =  13  Ibs. 

13 

Hence  P  =  — — -  =  167  Ibs.  Use  150  Ibs. 
0.0779  — — 

Air  pressure  curve: 

A  =  1.314 

U^   =   initial    air  vol.   =  188.7  cu.    in. 

paj    =   initial    air   pressure   =   1675   Ibs.   per   sq. 

*Ui  188.7 


Paf=Pai( ^'3  =  1675( >   l 

Ui-Ab  188.7-l.314b 

=  2731   Ibs.    per   sq.    in. 
where  b   =   length  of   recoil   in  inches. 

TABLE  OF  AIR  PRESSURES; 
A  188.7 


18 


O 


-in 

op 


fed 


\ 


J 


J§  1  I 
J.33J  Ml    pit  133 


T&    b£   U«       R!- 

I  I  I 


19 


RECOIL  CURVES 


X 


/ 


/ 


/ 


S 


tn 


20 


a 

b 

c 

a 

e 

f 

£ 

h 

182.  7 

X 

1.314X 

188.  7 

1 

logd 

1.  3*e 

l»j 

p 

1.314x 

188.  7 

1.675 

tf 

-1.314X 

0 

0 

188.  "70 

1.  000 

.00000 

3.  22401 

1675 

2 

2.63 

186.07 

1.014 

.  00604 

.00785 

3-23192 

l7o6 

6 

7.88 

180.  82 

1.043 

.01828 

.02376 

3.24777 

1*769 

14 

18.  40 

170.30 

1.  105 

.04336 

.05637 

3.  28038 

1907 

22 

26.91 

159.79 

1.  180 

.07188 

.093*4 

3.31745 

2077 

30 

39.  42 

149.  28 

1.  264 

.  1O175 

.  13228 

3-  35629 

2271 

38 

49.93 

138.  77 

1.359 

.13322 

.  17319 

3.39720 

2496 

42 

55.  15 

133.55 

1.  412 

.14983 

.  19478 

3.41879 

2623 

Throttling  Contraction  Areas: 

Ax 

x1  =  floating  piston  displacement  =  

Aa 
Constant  orifice  from  a  to  b  =  .0631  sq.  in. 

x1  =  9.685"    x  =  39.0" 

Variable  orifice  from  b  to  c  ;  at  b  =  .0631  sq.in. 
decreasing,   a*  c  =  .0302  sq.in.  uniformly 

x'bc  =  .0787"    xbc  =  0.4" 

Variable  orifice  from  c  to  d  at  c  =  .0302  sq.in.  de- 
creasing 

x<Id  =  .3958    xcd  =  1.6"    at  d  =  .0184  sq.in, 

uniformly 

Diagram  showing,  air  pressure,  throttling 
areas  and  pa  against  displacement  of  recoil. 
See  fig.  (6). 


Table  pf  p£ 


Recoil 

Pa 

A 

TP* 
*a 

.53Pa 

Rf 

*f 

^ 

Pa 

2 

1706 

1800 

850 

1270 

250 

2050 

6 

1760 

1860 

880 

1300 

260 

2120 

12 

1870 

1980 

930 

1350 

270 

2250 

20 

2027 

2145 

1010 

1430 

290 

2430 

28 

2210 

2340 

1100 

1520 

300 

2640 

52 

2305 

2440 

1150 

1570 

310 

2750 

36 

2416 

2550 

1210 

1630 

320 

2870 

40 

2533 

2700 

1270 

1690 

340 

3040 

44 

2665 

2820 

1330 

1750 

350 

3170 

46 

2730 

2890 

1360 

1780 

360 

3260 

Substituting  numerical  values  in  the  equation, 


ab 


=  sh_+ 


K*AaAaVa 


v. 
pi1,   we  have  p   =  140+.01  —  + 


1.067pa 

Substituting   in  the   equation  R  =    .184p+420.    we     have 
V* 


R=.  001845  —  +  0.1968  p£  +  443.  Substituting  R  in 

w°  the  equation  for 

the  total  resistance  to  recoil  K,  we  have,  at  the 
maximum  elevation  =  19°. 

K    465-57   11  +.043  p"  +  8.58.   Then  substituting 
* 


m 


10  • 


in  the  dynamic 
equation,  we  have  Vfm-Vf  n-(0  .116V*  +  .043pa'v8.58) 


22 


t  =  Vm-V,,  where  V*  = 


and 


t  =  xm-x, 


2  2 

On  the  bases  of  these  equations  the  following 
table  is  made  for  the  restrained  velocity  and  displace- 
ment during  the  powder  period  and  subsequent  retard- 
ation until  the  variable  orifice  at  the  end  of  re- 
coil is  reached. 

It  is  to  be  noted  that  after  the  powder  period 
(in  the  above  equation)  Vfm-Vfn  *  0. 

Retarded  velocity  curve  during  powder  period, 

El=  19° 
a  =  .116V* 
b=  .043P£ 

c   =    .116Vg+.043p£+8.58 


116    .043P" 
_y  a  + 

a        b 


.58 


Recoil  a 

b 

C 

Ct 

Vf 

Vt 

YI  t+V1 

t 

2t 

X 

2x 

1.33 

59 

•78 

146 

.29 

3.65 

27.52 

2^.84 

OO2 

007 

.  1105 

2.36 

78 

78 

165 

.49 

2.45 

29*48 

28.50 

003 

010 

.1963 

3.80 

94 

79 

182 

•  73 

1.77 

30.52 

3O.OO 

OO4 

014 

.3168 

4.90 

104 

80 

193 

.  58 

.72 

30.  66 

30.59 

003 

017 

.4081 

5.27 

108 

81 

198 

.  20 

.37 

30.83 

30.75 

001 

~18 

.4389 

5.64 

110 

81 

20O 

.  20 

.  21 

30.84 

30.84 

001 

019 

.4697 

Retarded  velocity  curve  during  retardation 
period  with  constant  orifice  Cl.=  18°. 


After  powder  pressure  (constant  orifice) 


23 


Fig.  7 


vr  +v 

Recoil 

a 

b 

C 

cd 

Vr 

S      1 

t 

2x 

2 

5.  64 
6.46 

110 
110 

81 
82 

200 
.  202 

.  2 

30.  84 
30.  44 

30.  84 
30.54 

.  001 

.001 

.4697 
.5376 

8.  61 

106 

82 

19*7 

.79 

29.25 

29.  65 

.  004 

.7167 

13.98 

96 

105 

190 

1.  9« 

26.52 

27.47 

.01 

1.  1635 

19.89 

•76 

90 

175 

1.75 

22.  94 

23.  82 

.01 

1.6573 

2*7.30 

66 

161 

161 

3.22 

18.  O4 

19.65 

.02 

2.  2713 

34.50 

4O 

144 

144 

2.  88 

12.  08 

13.  52 

.  02 

2.  8717 

39.00 

22 

134 

13* 

2.  68 

6.  64 

7.98 

.  02 

3.2453 

Substituting ,  as  before  the  value  of  —  in  t"he 

•_ 
dynamic  equation,  where  wo  is  now  wx  and 

variable,  we  have, 


466. V8 
V«-Vm  =  ._  .    +  858+  .043P£  t   and  substituting 

the  values  wv 


10  «w* 


obtained  in  the  proceeding  work,  we  have,  (fig.  7) 


Retarded  velocity  curve  during  the  retardation 

period  with  variable  orifice  towards  end  of  recoil, 
Cl.  =  19° 


466   V* 
a  =  —   -7  +  9.6 


b  =  .043 


Variable  Orifice. 


vr  +vf 

Recoil 

W0 

a 

b 

a+b 

(a+b)t 

Vr 

i 

t 

2x 

2 

39 

.  0631 

22 

112 

134 

2.  68 

6.64 

7.98 

.02 

3.2453 

40.  22 

.0295 

s7o 

28 

115 

143 

1.43 

3.92 

4.  64 

.01 

3.3517 

40.  84 

.0270 

228 

16 

117 

133 

1.33 

1.23 

1.  90 

.01 

3.4031 

40.95 

.  0263 

692 

12 

118 

130 

1.30 

.92 

.01 

3.4123 

CALCULATED  PRESSURES  AND 
PRESSURE  DROPS. 


The  total  resistance 
to  recoil  may  be  ex- 
pressed by  the 
dynamic  equation:  pA+E-Wrsin  (6-  K  where  p  =  sh0+Po 


PO  = 


K0Aa*A2V2 


paA     Rf    ab  Aa  Rf        BJJ 

Pf  -   Pa-Pa  »  — +-  -  P  'Pa-    ^  ~   1>P*I?  '  *T  P 

ii  A          A  SL  ol 


Rf  .01V8 

hence   Pf   =    .065  pa+-7    .0667(140+-—-   +1.0667p^ 
a  o 

Table   for  pressure   drop   across   floating   piston 
Rf  .01V« 

2.065pa+  ~r   .0667(140+  — ; — +! 
Ka     AJ  w2 


X 

.065pa 

R 
Aa' 

V2 

01V2 

ut  2 

O 

i.oep; 

.0667 

2 

109 

250 

762 

1820 

2190 

4250 

280 

70 

4 

113 

260 

912 

2300 

2240 

4680 

310 

80 

8 

117 

260 

918 

2320 

2320 

4780 

320 

90 

16 

126 

280 

681 

1720 

2500 

4300 

290 

120 

24 

137 

290 

445 

1120 

2700 

3960 

260 

170 

32 

149 

310 

240 

600 

2930 

3670 

240 

220 

40 

164 

340 

34 

414 

3250 

3800 

250 

250 

In  terms   of   the   intensity  of   pressure,   we 


have 


Wrsin0    R-Wrsin0 

r  :  nr—  •  —  — 


Let   a  = 


b  *    a+    Sh( 
c  =   b+pa 


*  =  c+pf 
e  = 


26 


Prom  these  values  the  following  table  was 
coEputed  for  the  various  pressure  drops. 


Re- 
coil 

RWrsin0 

A 

a+sh0 

b+Pa 

C  +  Pf 

d  +  P0 

2 

660 

800 

2610 

2580 

4630 

4 

720 

870 

2610 

2680 

5130 

6 

730 

880 

2540 

2720 

5270 

8 

720 

870 

2670 

2760 

5230 

12 

700 

850 

2720 

2830 

4980 

16 

670 

820 

2760 

2880 

4710 

20 

640 

790 

2820 

2950 

4450 

24 

620 

770 

2880 

3050 

4250 

28 

590 

740 

2950 

3160 

4070 

30 

590 

740 

3000 

3220 

4000 

32 

580 

730 

3040 

3040 

3910 

34 

570 

720 

3080 

3310 

3830 

36 

560 

710 

3130 

3360 

3750 

38 

560 

710 

3180 

3410 

3660 

39 

550 

700 

3210 

3430 

3610 

40 

530 

680 

3210 

3460 

3900 

41 

520 

670 

3240 

3480 

3480 

u"  «   travel   of   shot   in   the   bore 
ain_          2046*7.5     _    15340 
1.377*7.5      "    8.877 


n  =   b  +  n 


Vo   - 


1730x16.75 


1050 


a    7.5 

"•    1730 


max.   pressure 


89.87 
1730 

33.8 
27.6 

.0065 
36000   Ibs 


27 


Pb  =    1.12Pm  40300  Ibs. 

Pg  =  mean  constant  pressure  of  breech 

12WVm  15.95x1730 

=  =  14200 

64.4u"s          64.4x7.6x6.95 


_f/27   36000        ,.        /~      2736000.,        ,. 
89.87[( — x  -   i)  +  /(I  -  —  )«   -   1] 

*      IV16  14220  -  1614220' 


14.08 

27       14.08   x89. 87x40300 
PK  *  —  x  . 


(103.86)8 

4315 

2(Vg-V0Wr)   ^  2(33.8-27.6)1050      _   2x6. 2x1.050  , 

*•*   Pobxs"xg         "   4315x6.95x32.2  "  4.315x6.95x32.2 

.01248 
T   =   tt+tf    =    .0065+   0135  .02   seconds 

CALCULATION  AND  COMPARISOS  OP  AVERAGE  RE8I8TASCK. 

89.87x16.75 


12(Wr+W+if)  12x1067.53 


=    .00005   +    .372 

=    .372 

E    =    .1175   +    .372  =  .49 

T  *    .019  second 
E   =    .50'   =   6 

9 

-  mV|        -  32.6x33.8 

K  «J  -  »-4  -  =   5300    (Ibs) 
3.  4-.  5+33.  8x.  019 

-nVf        -  32.6   x  33.8 

K  =  2  —  L  =  £  -  =  5500    (Ibs) 
b  3.4 

Average    from   the   curve   of   forces   4230 


28 


A   =   1.314   hence   total    force   =  5530   Ibs. 


PEAK  VALUES 

Curve  of  resisting  forces  gives  an  average  5550 
Ibs.  for  K.   Peak  value  of  K  from  the  curves  5280 
x  1.314  *  6940  Ibs. 

6940 

Peak  ratio  d  =  =  1.25  with  respect  av.re- 

5530        coil  force  from  curve 

6940 
Peak  ratio  d'  =      =  1.30  with  respect  com- 

puted  av.  recoil 
reaction. 


RELATIVE  SIZE  OF  ORIFICE  UNDER  DIAPHRAGM 


AND  COUNTER  RECOIL  THROTTLING 


ORIFICE. 


Qf  =  Aaxf  =  5.303  xf 
Qh  =  Ax41=l.  314x41=53.  9 


5.303 
A  =  .3945 


•7087-. 6811  x  .35  =  ^^-   x  .35  =  .00346 
2.7953  2.7953 

dx  =  .00346+. 6811=. 6846  in. 

Ax  =  .366 

Smallest  area  under  the  diagram  .  3945  -  .366  =  .0285 

sq.in. 
Largest  throttling  area  =  .00665  sq.in. 


29 


.-.  the  throttling  take  place  not  under  the  diaphragm 

REIATIVE  DISPLACEMENT  OF  FLOATING  PISTOK. 

Aa  :A::xr:Xf  where  xr  =  displacement  of  recoiling 

parts 

Xf  =  displacement  of  floating 
piston 


xf 


1.314 
5.303 


Variable   throttling   begins   =f  =  5.3545   on  recoil 
corresponding  xr  =   5.3028x5.166  =  20<80 

1<314  take    .305   in. 


Total  floating   piston  displacement  =  Xf  • 

x41  =  10.1595 

Variable  orifice  ends  at  41-20.5=20.5  of  counter 

recoil 


THROTTLING  AREA. 


Dis- 

Depth 

Area 

Total 

Recoil 

Recoil 

tance 

of 

of 

Orifice 

Disp. 

in 

groove 

one 

inches 

gun 

5.  9055 
5.3150 

.0469 
.0453 

.003325 
.003212 

.  00665 
.00642 

2.38 

20.  50 

22.  88 

4.52*76 

.0421 

.002985 

.00597 

5.56 

26.06 

3.  7402 

.0386 

.002737 

.00547 

8.  74 

29.  24 

2.  9528 

.0339 

.  0024O4 

.00482 

11.92 

32.  42 

2.5591 

.030*7 

.002177 

.00435 

13.50 

34.OO 

2.  3622 

.028*7 

.  002035 

.0040*7 

14.30 

34.  80 

2.  1654 

.02*72 

.  00  1  9  2  8 

.00386 

15.09 

35.59 

1.  9685 

.0248 

.  C01758 

.00352 

15.89 

36.39 

1.7717 

.0220 

.  OO1560 

.00312 

16.68 

37.18 

1.  5*748 

.0193 

.  OO1368 

.00274 

17.48 

37.98 

1.3*780 

.0157 

.001113 

.00223 

18.27 

38.77 

.9843 

.0063 

.0004467 

.00089 

19.  82 

40.  32 

.  5906 

.0059 

.000413 

.00083 

21.45 

42.  05 

.  0 

30 


=  Pa  -  Floating  piston  friction 


Rf  =  .53  Pa  *  420 


C  'Recoil 

pa 

Rf 

Pi' 

0 

25*70 

340 

2,230 

i 

2530 

340 

2,  29O 

2 

2510 

330 

2,  160 

3 

2480 

330 

2,150 

6 

2390 

320 

2,0*70 

12 

2240 

300 

1,940 

20 

205O 

290 

1,*760 

28 

1890 

2*70 

1,  620 

34 

1*790 

260 

1,530 

3*7 

1»7<0 

26o 

1,480 

38 

1*730 

260 

1,4*70 

39 

1*720 

250 

1,4*70 

40 

1*710 

250 

1,  46O 

41 

1*700 

2^0 

1,45O 

VELOCITY  OP  COUNTER  RECOIL. 


=  AVr  where  V^  =  velocity  of  floating  piston 
Vr  -  velocity  of  recoiling  parts 


A  =  quantity  of  oil  throttled  = 


31 


K"a2 

P ' i  =  =  P  =  hydraulic  piston  cull  (where  PI i-Pa 

W2  a   a 

X  R. 


PA  -F=MrVf  

dx 

R  =  .1211  P+279+.0634P+138  =  .1845p+417 


w«      175W*  W* 


PA  -  .1845  P  -  417  =MrVr 


dvr 


P(A-.1845)-417=MrVr  — 

K'V2  dVr 

(P'i—  »T  -  -)(A-.1845)-417=MrV_  - 

r 


x  r  dx 

K'V2  ! 

(Pa1  ~  "P  —  )(A-.1845)dx-417  dx  =  5-  M(V2-V2) 

X  12 


AVgRAQB     VELOCIT?     OF     AK     IBTEGRAL     PfPIOD. 

In  computing  the  velocity  of  recoiling  parts 
by  the  formula 

V.+V 


Pa  A  dx  --  j;  --  R  dx  =  \  mr(V»-v»  ).  We 

Vi*V2 

can  take  for  value  of  (—  —  ^)2   as  it  simplifies 

2 

the  equation  to 

obtain  V^  .   It  is  nore  accurate  than  taking  either 
Vi  or  Va  as  the  velocity  f<>r  the  integral  period. 

Its  deviation  from  the  actual  average  is 
very  small.   Proof 

Let  V,  =  2  and  V2  =  2.25.   Actual  average  velocity 
of  the  intervals  V  V  = 

1      2 


32 


VJ  =  5.0625 

V;+VJ  9.0625 

Approximate  average  =  =  — — - -  =  4.5312 

2       2 

4.5312-4.5156         1.56 
4.5156 


COMPUTATIOK     OF     K. 

dv 

At   max.    velocity  —  =0 
dt 

pA   =   ZR   K^v» 

P=paf   "^T" 

x  K  V* 

ZR   =    .1845   p    +   417   =    1845(P^.    --  ~)    +   417 

W 


x 


K  V8  K  V2 

.'  •  PaA-  d—A  =.1845(P,  -  ~j- 


a 
x  x 


\  4  *'   *  /  \  f  «  *  —  y  •"" 

wx  A  wx          A 

417 

Kiv*        "T 

(Pj'   ""     ...j    '  ~    -  -  -  -      =   369 

*          i_    'I845 
A 

Assuming   max.    velocity  occurs   on   16"   of   counter   re- 
coil: 

1850-370    =  =    1480 

*x 

-1*V«  fK(4.9714-.7006+.3314)|i~|l2v2 

a  5.303 

=    1480  


175WJ  175   x    (.0065)8 


x 


33 


(Kx4. 6022x1. 314)2   3.28 
"      «    =  1480 

175   x    (.00665)      x    5.303 


1480xl75x.  00665      x   5.303 
K2    =•  -  ;  —   =   ?      K  =    .9048 

(4.6022x1.314x3.28) 

Use   K  =    0.9 

EQUATION  OF  FORCES  OK  COUNTER  RECOIL. 

Kiy>    [K'Ua-Ab  +  a^—V)*   [^( 

"a 


W«  175W*  175    x   w» 

(1.11x4.  6022x1.  314)2V*   _    9159     V^ 
(13.24x5.303)8W»  106        W^ 


dx-  417  dx 

10345     V« 
(1.1295   Pit   -  -T-TT-     rjj 

J.U  "X 

(.06929P£i   -  634>66?6     —  -  25.5828)dx=V?-V? 


.     317.33dxN   ,  , _      317.33, 
Vl          10eWa — }=   *  U"   10*W«        + 

X  X 

For  variable   orifice. 


For  constant   orifice 

V*  (1-f  .5979dx")=V«(l-.5979dx")-»  (.005774?^  i-2.1316)dx" 

t  i  * 


7 


i 


2 


Ill 


35 


36 


II 

PI  <M 


c-     o 

10       rl 


PL, 


c 

OS 

•»» 

M 

C 

o 

V 

XX) 

C 


P 


-a 

H 
0 


X 

TJ 


0 


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LOG/.PITHMIC     METHOD 

(Check  on  velocity  of  c'recoil) 

cV*     dV 
F-,-Wr(sin0-n  cos  0)-  ZR =  roV  — 

W>      dx 

cV2  _     dV 

x 
cV2     .  dV 


W  dx 


A=1.1295   P^i   -   417 

10345 
1C6 

cV*  cV*          2c(x2-xt) 

log(A-—  )-    log    (A-—  )-2>3mrW. 


cV*  cV*  2c(x-x) 

T~  -  A)   -   log(—i  -   A)   = 


W»  '    W£  2.3   m«« 


°g 


cV2  2.3m   W* 

(  —  -  -   A) 
wa 
"x 


2c(xf-xl) 
2.3m   W* 


40 


•H-i 

°v§  w$ *     

^~ " As 

2c  .-*.> 


Antlg   — 


2.      m  W« 


V|   =    t  -       +    A    1   — 
2c(x-x) 


2i 
Antlg 

2.3   mW* 


2CCX.-X.  )  _   2x10345  dx      _   276  dx 

2.3mW*  "   2.3xlO«x32.6  W»      =   10^  W' 


.. 


.      276  dx  «£. 

Antlg         r|          w* 


A  =   1.1295   P^«   -  417 

10345          C 
x 

276        dx 
a  =  " 

10 6        W| 
b    =   Antlg    a 


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63,300 

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10,070 

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640 

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13440 

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C.R. 

V 

V2 

W*10« 

cV* 

W2 
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p/i 

P 

1845] 

ZR 

ZR 
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sq. 
in. 

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2570 

2230 

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410 

830 

630 

1600 

1.90 

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44.  2O 

750 

2550 

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8.21 

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1700 

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2180 

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90 

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390 

90 

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8.50 

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2480 

2150 

380 

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490 

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2.93 

8.51 

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1730 

2420 

2100 

370 

7o 

490 

37o 

10 

7 

2.  89 

8.35 

44.  2O 

1690 

2360 

2040 

350 

60 

480 

37o 

20 

10 

2.84 

8.01 

44.  2O 

1630 

2280 

1920 

290 

50 

47O 

360 

60 

15 

2.  73 

7.09 

44.  20 

1530 

216O 

1850 

320 

50 

470 

360 

60 

20 

2.  57 

6.60 

44.20 

1430 

204O 

1750 

320 

50 

470 

360 

60 

23 

2.  42 

5.83 

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1380 

1980 

1670 

290 

50 

47O 

360 

60 

26 

2.  23 

4.96 

33.99 

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1930 

1640 

300 

50 

470 

360 

60 

29 

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1310 

1875 

1580 

270 

50 

470 

360 

60 

31 

1.  83 

3.34 

24.  Ol 

127O 

1840 

1560 

290 

50 

470 

360 

60 

33 

1.62 

2.  62 

19.01 

1260 

160O 

1530 

270 

50 

470 

360 

60 

35 

1.33 

1.79 

13.32 

1350 

1780 

15OO 

300 

50 

470 

360 

60 

1 

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R 

:'  Re- 

•  V 

V2 

W2  

pa 

Pa' 

P 

3845p 

R 

0 

coil 

e 

10* 

X 

1. 

36 

1.  15 

1.32 

10.  24 

1180 

I76o 

148O 

300 

57 

474 

361 

60 

3*7 

.96 

.92. 

7.29 

1150 

1750 

148O 

330 

61 

478 

364 

34 

38 

.72 

.  52 

4.  2O 

1130 

174O 

14*70 

340 

63 

480 

365 

25 

39 

.35 

.12 

1.  OO 

1100 

1720 

146O 

360 

66 

483 

368 

8 

40 

.  29 

.  095 

.72 

1080 

1710 

1455 

375 

69 

486 

370 

5 

41 

.  25 

.  08 

.70 

1050 

I7oo 

1450 

400 

74 

491 

374 

26 

DESIGN  LAYOUT  OF  COUHTER  RECOIL. 

Divide  the  total  length  of  recoil  into  four  parts: 
LQ  =  from  0  velocity  of  counter  recoil  to  max. 
velocity   fa) 


LO-LQ 


Lo  =  one-half  of  counter  recoil 


distance   (b) 
£  =  last  half  of  counter  recoil  distance  -  6  inches 

(c) 
=  last  6  inches  of  counter  recoil  (d) 


44 


cV1          dv 

a   wx  dx 

(a)     Assume  a  max.  velocity  of  2.95  ft/sec. 

When  velocity  is  max.  V  -^  =  0 

dx 

cV* 

.'  .  (Pi )  A-  2R=0 

W0 

P.  -£ll  -  5 

a    W2 
wo 

cV2 
2130 =  370 


175  x  W* 


Wx 

(i  x4. 602x1. 314*2.  93)2 

_9 _  W2 

(5. 303x13. 23x41. 95)2 
WQ  =.00669  sq.in.  from  drawings  WQ  -  .00665 

Deviation  .45* 

(b) Orifice  is  constant  during  this  period 

W0  =  .00665 

cV2        dv 

(P •  -  — -)A-2F=mV  -—   equation  of  forces  with  one 
W          dx 

unknown  velocity 

Find  V  at  "Lo" 

(c)     Calculate  the  unbalanced  force  at  the 
beginning  of  the  interval.  This  force  is  constant 
up  to  six  inches  from  the  bettery  position. 

cV2          dv 
(Pi  -  - — )A-2R  =  mV  -— 

x      i    :"vV 

mV  dv  =  f6  dx  —K  " 


V  - 


32.6      32.6 


f..  =  /V2  -  .  307 

A 


45 


C'Re- 
coil 

x" 

.3x 

V«-.3xn 

vr 

Vr 

diff  . 

H 

20.  5 

0 

0 

6.44 

2.54 

2.54 

00 

23 

2.5 

.n 

5.  64 

2.  38 

2.42 

.  O4 

1.65 

26 

5.5 

1.6-7 

4.  "77 

2.19 

2.  23 

.  04 

1.79 

29 

8.5 

2.  60 

3.  84 

1.96 

2.  00 

.  04 

2.  00 

31 

10.5 

3.  22 

3.  22 

1.  8O 

1.83 

.03 

1.66 

33 

12.5 

3.83 

2.  61 

1.61 

1.62 

.01 

.  62 

35 

14.5 

4.5 

1.  92 

1.3*7 

1.35 

.  02 

1.43 

36 

15.5 

1.  14 

1.  15 

.  01 

.8*7 

(d)  Velocity   at   the  beginning   of    the   period  =   V  =1.14 


-V 


if 


2L 
V« 


m  Lra 


'    (—  -  2x) 
1     m    IK 


V?   - 

m 


=   82 


72    =  v,2   -  -    (2x  -  7,)=  V«  -  2.51(2x  -   24  x2) 
1 


46 


V2   -  2.51(7   *"-    -0167  x*>    =  V*   -  2. 51 (.167-. 0167) 

1  g  * 

=  V*   -.39 


C.Rj 

X 

V2 

design 

drawing 

V  -V 

^Deviate 

V± 

y 

36 

1 

1.31 

1.  15 

1.15 

o 

0 

37 

1 

.92 

.96 

.96 

0 

0 

38 

1 

.53 

.73 

.72 

.01 

1,37 

39 

1 

.  14 

.37 

.35 

.  02 

5.  40 

40 

1 

0 

o 

.09 

.09 

41 

1 

0 

0 

.08 

.  08 

Assuming   orifice   curve    a  parabola 


y2=Kx 


X 

*X 

WX 

Orifice 
by  para- 
bola 

Orifices 
drawing 

a-b 

^Deviation 

1 

8.  8< 

2.98 

.  0014*7 

.00125 

.  OOO  22 

16 

2 

17.65 

4.  20 

.00210 

.00206 

.00004 

2 

4 

35.  3« 

5.95 

.  00297 

.  00320 

.  OOO23 

7 

6 

53.  O1? 

7.  30 

.  00365 

.  OO410 

.  OOO45 

11 

8 

70.  7€ 

8.41 

.  OO420 

.00465 

.OO045 

10 

10 

88.  45 

9.45 

.  0047O 

.  00513 

.  00043 

8 

3  ? 

106.  13 

10.  30 

.  00515 

.00550 

.  00045 

8 

14 

123.82 

11.  11 

.  00555 

.OO583 

.  OOO28 

5 

16 

141.  51 

11.  89 

.  00594 

.  00612 

.  OOO18 

3 

18 

159.  20 

12.  6l 

.  00630 

.  00640 

.  OO010 

1.5 

20 

176.89 

13.30 

.  00665 

.  00665 

.  00000 

0 

I  I 


£ 

I 


W) 
L 


RECOIL  SYSTEM  WITH  VARIABLE  RBCOIL- 
IHPEPEHDBNT  RECUPERATOR. 

In  a  mechanism  of  the  type  now  to  be  discussed 
the  brake  cylinder  contains  a  buffer  rod  fixed  to 

the  recoil  cylinder  with  grooves  of  varying  depth 
for  throttling  during  the  main  recoil  through  the 
recoil  piston.   At  the  end  of  the  buffer  rod  is  a 
spear  buffer  which  enters  towards  the  end  of 
counter  recoil  a  smaller  buffer  chamber  in  the  pis- 
ton rod,  and  brings  the  recoiling  mass  to  rest  at 
the  end  of  counter  recoil.   The  recuperator  con- 
sists of  two  cylinders,  the  recuperator  cylinder 
communicating  through  a  passageway  of  suitable 
cross  section  to  the  air  cylinder.   The  air  cylinder 
contains  a  floating  piston  which  separates  the  oil 
and  air.   A  valve  controlled  by  a  weak  spring  in 
the  oil  end  of  the  air  cylinder  opens  wide  during 
the  recoil  and  thus  the  throttling  through  it  during 
the  recoil  is  negligible.   In  the  counter  recoil, the 
valve  is  closed  by  its  spring  and  the  return  flow 
of  the  oil,  and  throttling  takes  place  through  small 
holes  in  the  valve,  thus  lowering  the  pressure  con- 
siderably below  that  of  the  air  in  the  recuperator 
cylinder.   During  a  greater  part  of  the  counter  re- 
coil, the  air  pressure  in  the  recuperator  is  just 
sufficient  to  balance  the  total  friction.   Towards 
the  end  of  counter  recoil,  the  recoiling  mass  is 
brought  to  rest  by  the  reaction  of  the  buffer  in  the 
hydraulic  braking  cylinder. 

EXAMPLE  AND  CALCULATION.     As  an  example  of  a  suc- 
cessful model,  a  com- 
plete calculation  will  be  given. 

GEMERAL  DATA. 

A|J  =  area  of  bore  29.2  sq.in. 

w  =  weight  of  projectile  96.1  Ibs. 

wr  =  weight  of  recoiling  parts  9124  Ibs. 

w  =  weight  of  powder  charge  26  Ibs. 


u  = 

Pbm 


b  = 

0  - 

v  = 

S 
v 

Ph 

Pai 

Paf 
Vi 
Vf 
A  = 


49 


travel  of  shot  in  bore  185.68  (in.) 

=  max.  powder  pressure  on  breech  31,500  Ibs/ 

sq.in, 
height  of  center  of  gravity  of  Wr  from 

ground  5.7  ft. 

length   of    recoil  70.78   ft. 

angle   of   elevation  Oc 

velocity  of  projectile   in  bore 

=   volume   of   powder  chamber  1334   cu.in. 

=   hydraulic   pressure  Ibs/sq.in. 

=    total   hydraulic   pull  Ibs. 

=    initial    air   pressure  Ibs/sq.in. 

=   final    air   pressure  Ibs/sq.in. 

=   initial    air   volume  cu.    in. 

•    final    air   volume  cu.    in. 

effective    area  of    recoil    piston 

190 
0.7854 


25.4 


25.4 


=  0.7854(56.955-18.755) 


29.22  sq.in. 


Ar  =  effective  area  of  recuperator  piston 

82   a     40 .* 

'25.4'     25.4 

=  0.7854(10.427-2.48)  6.24  sq.in. 

Aa  =  area  of  air  cylinder  or  floating  piston 

137 
=  0.7854( )  22.85  sq.in. 

25.4 


Free  velocity  curve 


By  Le  Due's  formula 

Velocity  of  projectile  in  bore  V  =  where 

u"  b+u 

u  =  — -  =  travel  of  projectile  in  bore  -  in  ft. 

J  C 


50 


a   =   6823  x  A 


A   = 


w*27.68 


a  = 
log 


26x27.68 
1334 


.5395 


6823X.5395 

vx.c 

a  =   log    6823    +  ~      (log   5395-   log   1000)+  Jj 

26-  log    96.1)   =  3.83398+  —(3.73199-4.00000) 
+  7(1. 41497-1, 98272)   =  3.83   398  +  . 31100-. 33333 


+. 707485- . 99136  =  3.52778 


a  ^  3370 


4W        1.12a2    _   4x91.1x1.12x3371 
"27   X    gAgPm      "27x32.2x29.2247x31500 
3371x154733  52161 


=  6.1124- 


=  2416  ft/sec. 


6.1124+15.4733 


21.5857 


Vf   =   free   velocity  of   recoiling   parts   = 


Vn(W+.5») 


_     P 


=  free  displacement  = 


W, 


=    .01196   V 

•=    .01288u 


u 

u 

au 

b  +  u 

V  —  j_ 

.  01196? 

.01288V 

in. 

1 

12 

3370 

7.  1124 

474 

5.67 

.  155 

2 

24 

6740 

8.  1224 

832 

9.95 

.  294 

3 

36 

10110 

9.  1124 

1310 

13.  28 

.464 

4 

48 

13460 

10.  1124 

1330 

15.91 

.  618 

5 

60 

16820 

11.  1124 

1512 

17.99 

.774 

6 

72 

20200 

12.  1124 

1667 

19.  84 

.928 

8 

96 

26900 

14.  1124 

1900 

12.  61 

1.  235 

10 

120 

33700 

16.  1124 

2090 

24.  87 

1.545 

12 

144 

40400 

18.  1124 

2230 

26.  54 

1.  854 

13 

156 

43800 

19.  1124 

2290 

27.  25 

2.  010 

14 

168 

4»720O 

20.  1124 

2340 

27.  85 

2.  l6l 

15 

180 

50550 

21.  1124 

2390 

28.  44 

2.32 

15.473 

186 

52150 

21.  5854 

241C 

28.  69 

2.392 

51 


CALCULATION  OF  S  ANP  T. 


'=t0+tlQ  Bhere  to  =  time,  shot  to  muzzle 

t   =  time,  gas  expansion 
*o 

a  u    »    16.473 
V   t  f  "  *  '   2416  "  '°°  63  seconds 


0  ~  —————  where  VfjB  =  velocity  of  free 
P0t   S°  g  recoil,  mas. 

VQ  *  velocity  of  free 
recoil  when  shot 
is  leaving  muzzle 
Wf  =  weight  of  recoil- 

ing parts 

PQk  =  muzzle  pressure 
at  base  of  breech 

S0'  =  cross  section  of 

the  bore  in  sq.  in. 

W  V-+4700  w 

=    •  _  96.1x2416*4700x26           . 

max   -  =  -  =  38.78  ft/sec. 

W  9124 

2416(96.140.5x26) 

=  28.83  ft/sec. 


wr 

max.  pressure  on  breech  =  1.12  Pm  =  35,300  Ibs/ 

sq.  in. 
mean  coast  pressure  on  breech 

W  V2 

-    =   19200   Ibs/sq.in. 


64.4 


.   .   ..,  (S  !l  -  I).  /!   -  *1  !•)'   -   1    ]    =   67.38 
16  Pe  16   Pe 

27  u"  27       r--~c        185.68x35300 

p       *  —  e*   — — —      PV   •—  x   57.38     x   • 
ob          4          (e+u")»  4  (57375+185.68)* 

=  10141  Ibs/sq.in. 


pob*Aa  x  8 


2(38.78-28.83)9124 
10.141*29.225*32.2 


^    .01899 


T   =   i0    +   tt      =    .009631 +.01 899=    .0286   seconds. 

E   =   Xf 0+Xf  iQ   =   Sf   while    shot   travels    to  nuzzle 
+   Sf   gas   cap. period. 


w+.05w 


96.1+13 


)=.1826   ft. 


+vo*«.0  - 


32. 2*29. 22x1 01 4Qx. pi 899 
3.X9124  x 


+   28.83   x    .01899 
=    .5476   ft. 

E   =    .1826   +    .5476  =    .7308 
t0   =  |(2.3    log  Z*      +  Jl   +  2.098) 

-c  D       D 

6>1124'    (2.3  log  0.327  u  +   "  •   +2.098) 
3370  6.1124 


t0  =  .00414  log  0.327  u  +  .000299  n  +  .00378 

time  up  to  (shot  to  muzzle) 
w 

^  *   f~  f\  '          ^ 

i   =  —————   —  =  will  give  time  correspond- 

^O     P   x^"xt$        a 

rob  ^a  *         ing  to  Vf  during  gas  ex- 
pansion period 


0 
to  =  - 

.00412      0.0277  u"    0.0000249  u" 

0.00378 

BAG 

Xf 

q" 

0.0277u" 

log  A    B 

c 

t0 

•  354 

36 

i. 

0 

0 

.0009864 

.  00468 

.566 

48 

1.3296 

.  12385 

.  OO0513 

.O011952 

.00549 

.920 

78 

2.  1606 

.33465 

.001385 

.001942 

.  00*710 

1.345 

114 

3.  15*78 

.  49941 

.002068 

.002839 

.00869 

1.  628 

138 

3.  8226 

.58240 

.002411 

.  003436 

.  00963 

1.  840 

156 

4.  3212 

.63558 

.002631 

.003884 

.01049 

2.053 

m 

4.  8198 

.  68305 

.  OO2828 

.004333 

.01094 

2.  191 

185.  65 

5.1425 

.  *71122 

.  OO2944 

.  OO4623 

.01134 

53 


2W, 


(Vf-V0)=  ,001912(Vf-V0) 


Vf 

V0 

Vf-V0 

*.„ 

t0 

3.  00 

31.  30 

2.  47 

.00472 

.01606 

4.OO 

33.  66 

28.  33 

4.83 

.  O0924 

.02058 

5-  OO 

35.40 

6.57 

.  01256 

.02390 

6.00 

36.85 

8.02 

.01534 

.  02668 

"7.00 

3*7.81 

8.98 

.01717 

.02851 

8.00 

38.55 

9.72 

.01859 

,02993 

6.  76 

38.  68 

9.85 

.01883 

.0301*7 

Resistance   to   recoil. 


^ 


Slope  of  the  stability  curve  =  -  =  R  -  r  where 

b  =  length  of  recoil 

h  =  height  of  center  of  gravity  of  recoiling 

parts  from  the  ground  (5.70) 
Wr  =  weight  of  the  recoiling  parts. 

R  =  max.  resistance  to  recoil  in  battery 
_   r  =  max.  resistance  to  recoil  out  of  battery 

Wrb    9124*70.78 


R-r=- 


=    9450 


h  68.4 

Total  variable   resistance   to  recoil  =  K  = 


*  — 

max        2b 


R-r 
b-E*VfT-  - —  (b-E) 


Tg 


I 

•HrH 


K  = 


»  t 

x    38.78  +    800(5.9058-. 730) 


5  .  9058-  .  730+3  8  .  78*  .  0286-1  600  (5  .  9058-  .  730  ) 


(.0286)' 

2  — 
32,2 


K  =  37200   Ibs.    in  battery. 


54 


R— r      KT8 

K1  out  battery  «  K  -  :( )  (b-E+ )  =37200-1600(5.9058 

2Mr 

-.730+. 0538) 

K1  =  29600  Ibs.  out  battery 
Construct  a  curve  showing  K  at  each  point  of  recoil. 

OALCPHTIOJ     OF     AIR     PRESSURE. 

A  =  effective  area  of  recuperator  piston 
V^  =  initial  air  volume   1620  sq.in. 
Vf  =  final  air  volume  =  V^-AX 

pai  =  initial  air  pressure  =  1420  Ibs/sq.in. 


af 


final  air  pressure  =  P«i  (  —  ) 


1420 


1620 


1  .9 


1620-6.25 
log  Paf=  log  1420+1. 3[log  1620-log  (1620-6. 25x)] 

log  Paf  =  log  1420+1. 3[log  1620—  og (1620-6. 26  x)] 
Paf  =  1420  log  1.3  e 


Re- 

6.  25 

1620-6^ 

c  leg 

P.f 

o  '  r  e- 

o  e  1 

1   X 

I 

1.13* 

ooil 

X 

log  o 

e 

1.3* 

1 

6 

1614 

3.  20790 

.OO162 

.00211 

1.0049 

143O 

7o.7e 

f> 

31 

1589 

3.20112 

.  OO84O 

.01092 

1.0255 

1460 

65.78 

10 

63 

1557 

3.  19229 

.01723 

.02240 

1.0529 

150O 

6o.7s 

15 

93 

1527 

3.  18384 

.  O2568 

.03338 

1.0799 

1530 

55.78 

20 

124 

1496 

3.17493 

.03459 

.04497 

1.  1090 

1580 

6o.78 

25 

155 

1465 

3.  16584 

.04368 

.05678 

1.  1405 

1620 

45.78 

30 

186 

1434 

3.  15655 

.05297 

.  06886 

1.  1750 

1670 

40.  78 

35 

217 

1403 

3.  147  06 

.  06246 

.08120 

1.  2055 

1710 

35.78 

40 

248 

1 

372 

3.13735 

.07217 

.09382 

1.  244O 

I77o 

30.78 

45 

279 

1344 

3.12743 

.  08209 

.  10672 

1.  2788 

1820 

25.78 

50 

313 

1 

307 

3.  11628 

.  O9324 

.  12121 

1.  322O 

188O 

20.78 

55 

341 

1279 

3.  10687 

.  10265 

.  13345 

1.3595 

1930 

15.78 

60 

375 

1245 

3.09517 

.11435 

.  14866 

1.  4080 

2OOO 

10.78 

65 

406 

1214 

3.08422 

.  12530 

.16289 

1.4550 

2  >7o 

5.78 

68 

425 

1195 

3.  0»7I73'7 

.  1J215 

.17180 

1.4850 

211-> 

2.78 

"70 

438 

1182 

3.  0"7262 

.  13690 

.1*7  179*7 

1.5  )6g 

2140 

.78 

•70.    78 

439 

1181 

3.  07225 

.  13*729 

.1*7845 

1.5^80 

214O 

o 

Velocity  of  retarded  recoil. 
During  T 

Vr=Vf  —  — -t      where  K  =  retarding  or  breaking 
force  constant,  during  T.  = 
37800  Ibs. 
Vr  -  Vf  = 

37200x32.2 

-  t  =  Vf  =  131.28  t 


v 

Xr=Xf    =  — — -   t2    where  Xr   =  retarded  displacement 

&M*» 

37200x32.2  66_ 

x9124 


t 

t2 

*f 

65.64t* 

*r 

Vf 

131.  28f 

Vr 

.  01085 

.  000122 

2.  19 

.008 

2.  1 

28.83 

1.32 

27.51 

.  0128 

.000164 

2.954 

.  oio75 

2.  83 

31.5 

l.6e 

29.82 

.  0148 

.  00022 

4.  "734 

.  01445 

4.  56 

35. 

1.94 

33.06 

.  02984 

.  00089 

8.76 

.  0585 

8.06 

38.78 

3.91 

34.87 

After  T, 


R-r 


K  =  K  +  (b-x)   =  29600*1600(5. 9058-x)  =  39  060 

A        b 


-1600  x 


-J 


V«  - 


37200+  K,  (x-. 737) 


283 


56 


x" 

X1 

Kx 

K**X 

nr 

X-.73 

J 

*5 

vr 

9 

.75 

36860 

265 

.02 

5.  30 

1215 

34.  9< 

19 

1.  585 

36520 

261 

.  855 

223 

99*7 

31.6 

29 

2.418 

35200 

256 

1.  688 

432 

"788 

28.  0 

39 

3.25 

33860 

251 

2.52 

633 

587 

24.  2 

49 

4.O8 

32530 

246 

3.35 

825 

395 

19.9 

59 

4.91 

33  21O 

242 

4.  18 

1010 

210 

14.5 

69 

5.75 

29860 

23*7 

5.02 

1190 

30 

5.4€ 

•70.  "78 

5.9 

29610 

236 

5.17 

122O 

0 

0 

C  A  V, 


13.22 


C  =  1.39 


A  =  29.22 


*r 

Vr 

Ph 

v^ 

13.22/P^" 

CAVr 

w 

2.  1 

27.51 

955 

30.9 

408 

1110 

2.72 

2.  825 

29.  82 

950 

30.  8 

407 

12OO 

2.95 

4.  561 

33.06 

9*5 

30.78 

406 

1330 

3.  28 

8.  O6 

34.87 

940 

30.  62 

405 

1405 

3.47 

9. 

34.90 

936 

30.  60 

404 

1410 

3.49 

19. 

31.  60 

885 

29.  78 

393 

1275 

3.  24 

29. 

28. 

815 

28.  50 

377 

1130 

3. 

39. 

24.  2 

738 

27.  2 

360 

975 

2.71 

49. 

19.9 

662 

25.  8 

340 

8O3 

2.36 

59. 

14.5 

577 

24 

318 

585 

1.74 

69. 

5.4 

492 

22.  2 

394 

221 

.752 

•70.  87 

o 

477 

21.82 

289 

0 

0 

As  the  total  oil  displaced  by  the  recoil 
piston  does  not  pass  through  the  recoil  throttling 
grooves,  a  part  passing  into  the  buffer  chamber 
the  above  throttling  formula,  is  not  exact  and 

becomes  C(AVr-Qb) 

w  =  _____ — —   where  Qb=  the   quantity  of 
13.22/Pn  oil   passing    into 

the   buffer   chamber. 
With  this  value  of  w.     C 

should  be  given   a  value  of   about      1.5. 


57 


FRICTION     CAI/CULATIOH. 


Floating  piston  friction  = 


-a— 


:137&* 

a 


.069 


o 


^r 
in 
ii 
TS 


^37&-*.l37S*~ 


66CX)7 


Rr  =  K  A  P  f 


P  = 


air  pressure   (ibs/sq.in.) 

belv.  spring  pressure  42F=  load  7700  Ibs. 

7700 


.7854(5.38  -  4.37  ) 


=  1000  Ibs/sq.in. 


P£  =  belv.    spring    load   42E=   6600   lbs.=  860   Ibs/sq.in. 

K  =  poisson  ratio 

f   =  coefficient   of   friction 

A  =   area  of   frictional    surface 

Pt  =    intensity  of   pressure   1  to  A 

Rf  =  «d[(bf  +  af  »)Pt+   7   af'Pfl]K+   ivdl(bf  +  af  '  )P'+7af'  Ptf  ]K 

A  w    z        5  ^25 

d  =  5.4  in. 

b  =  .197  in. 

a  =  .1376  in. 

K  =  poisson  ratio  =  .73 

Pt  =  air  pressure  +  blv.  spring  pressure  (Pa+Po) 

=  pa  +  -  1000  Ibs/sq.in. 
a   3 

P'  =  oil  pressure  +  blv.  spring  pressure  (P^*P0) 


=  P, 


I     86° 


=  spiral  spring  pressure  +  P&- 


—  a* 


.7854x5.4 


58 


=   130  +   Pa      Ibs/sq.in. 
Assume  Pa=Pa 

Rf   =   3.1416x5.4[(.197x.05+.1376x.09)(Pa+668)+.069 

• 

*.°9(130+Pa)]. 73+3. 1416x5. 4[(.197x.05x. 1376 


x.09)CPa+573)+.069x.09(130+Pa)].73   =   16. 95[ ( .00986 
+  .01238)(Pa+668)+.0062(130+Pa)]    x    .73    +   36.95 
L  ( .00986+  .01238)  (Pa+573  )+  .  0062  (130+Pa  )]  x  .  73 


12.38(.02224Pa+l48.4+.  8     .0062Pa)+    12  .38C.02224P. 


+127+.8+.0062P_) 

• 


12.38(.02844Pa+l48.4+.02844Pa+127) 
Rf   =    .352Pa+3430      Ibs. 


RBCUPIRATOP  PISTOK  HEAD  FRICTIOK  «  R 


.1375 

J375J 

/ 

a 
lit 

a 

b 

f 

o 

Sf  ^NO    FRICTiON 

2    e)60* 

"  -•      ^ 

Z37 

< 

i 

Rt    -   K  A  P   f    =   «d(af  f+bf  )(P+P0)K  =   tr  »   3.225C.1375 
x.09+.237x.05)(980+P).73   =   10.14* 
.0242(980+p).73 


0.18  P0+175    Ibs. 


59 


RECUPERATOR  PISTON  ROD  PACKING  FRICTION 


.1375 


J 
l 


6160 


.156 


1 

-a 


R3    =   rr  x   1.57(.1375x.09+.158*.05)(860+p    6   ). 

=   4. 94x. 0203(860    +  p).73  =   63*. 073   p 
R3    =   60+. 073   pa        Ibs, 

RECOIL     BRAKE     STUFFING     BOX    TRICTIOH    «H. 


.1375 

,1375 

J375m 

a 

a  * 

1  a  ' 

\ 

} 

^  ' 

b 

K  ^ 

60* 

i 

AS 

4-/D' 

ii 
•o 

A      D   =    .7854(32.85-19.4)   =   10.6 


4150 

30.6 
R      =   n   x   4.4K-    .1375*.09+.158x.05)(394+Pe).73 

4-2  o 


60 


=    10. IK. 0186  +.0079)(   394+pg) 
R4  =    105  +    .268  ph        Ibs. 


GUIDE     FRICTION    -     R 


Rg  =  .15Wr=  9124*. 15=  1370  Ibs. 
R   =  1370 


FORCES  ACTING  CORING  RECOIL 

B  =  total  braking  pull  =  Ph+Pa 

P^  =  hydraulic  pull 

Pa  =  load  on  air  recuperator  pull 


Pa 


po 


=*=*= 


ARE1A  Of  PISTON    Au     II 


Fig.  A. 


Rg  GUIDE 
FRICTION 


I 

p'  =  (Fa  +  —  )  =  intensity  of  recuperator  pres- 

sure, Ibs/sq.in. 
R+  =  friction  =  R  +F  +R,+Rg 

T.  2346 


+R 


Rf  =  0.18 

R*  =  .073Pa+60 


R4  =  .268Ph+105 
Rg  =  1370 
2R=Rt  -  .253Pa+.268Pn=  1710 


I 


Ibs. 


61 


p 

Ph  - 


A+.268 


1710 


26.268          26.268 


22.85x26.268   26.268 
Rf 


«    .0381   p  -   0.2472   pa  -    .2376  --   65 

22.9 
but  Rt   =   .352   pa+3430 

Rf 

—-  =    .0154  pa  +    150        hence   ph   =    .0381    p-    0.2626  pa 


-  215    (Ibs/sq.in) 


P  =   K+Wr   sin  ft 
Ph   =    .0381   P  -    .2626  Pa  -  215 


Re- 

P 

.0381P 

Pa 

.2626Pa 

Ph 

soil 

i 

37200 

1417 

1430 

376 

826 

5 

37200 

1417 

1460 

383 

819 

10 

37ooo 

1410 

1500 

394 

801 

15 

36400 

1387 

1530 

401 

771 

20 

35800 

1364 

1580 

415 

734 

25 

35200 

1341 

1620 

425 

7oi 

30 

34600 

1318 

I67o 

439 

664 

35 

34000 

1295 

1710 

450 

630 

40 

33500 

1276 

1770 

465 

596 

45 

32900 

1253 

1820 

478 

560 

50 

32300 

1231 

1880 

494 

522 

55 

3l7oo 

1208 

1930 

506 

487 

60 

31200 

1189 

2OOO 

525 

449 

62 


65 

30600 

1166 

2O70 

544 

407 

6s 

30200 

1152 

2110 

555 

382 

70 

30000 

1143 

2140 

563 

366 

7b.  7fi 

29600 

1128 

2140 

563 

352 

Rg    = 
ZF  = 


— fa    *   975) 
.180  pa  +    175 

.073   pa+    60 

105+.268   p    h 

1370 
.263   pa    *    .268  p    h    +   2685 


Re- 
coil 

.263pa 

.268ph 

R 

phA 

APa 

5 

380 

540 

3605 

21250 

9130 

15 

400 

540 

3625 

20000 

9550 

25 

425 

530 

3640 

18200 

10110 

35 

45 

450 
478 

530 
520 

3665 
3683 

16350 
1455C 

10700 

138OO 

55 

510 

520 

3715 

12650 

12050 

65 

545 

520 

3750 

1060O 

12920 

10 

560 

510 

3755 

9530 

13380 

70.  81? 

560 

510 

3750 

9160 

13380 

COUNTER  RECOIL  CALCULATIONS. 

Oil    Ports 

Recuperator   valve    oil    ports:    see   fig.    (A)    Page   59, 
Ports    in   58B1 


=   2*.  7854    x 


•  *x 


=    .03896   sq.in 


63 


m 
m 
^~ 


m 


92'2 


HI 


<fr 
o* 


oO 


< 


in 


uJ 

L_ 

cQ 


«0 
1^ 


64 


BOFJBF  GROOVE, 


ro 
<t 

a 


,1781 


1594 


1407 


1220 


1033 


0546 


.0659 


0472 


0285 


0098 


=  .2362Xd 


=.0465 


=.0421 


=.0377 


=  .0332 


=  .0286 


.0244 


=  .0199 


=  .0156 


=  .0111 


=  .0067 


.00231 


65 


Recoil  brake  :-  filling  ir  buffer  port  holes 
Port  in  piece  29A 

aos*o  diameters  15. mm. 

a0  =  2  x  .7854(~)*  =  .5478  sq.in. 

20 

a  «  .7854  — •         .4869  sq.in. 
O         24 


Smallest  oil  port  in  counter  recoil  is  Wo 

;fore  W0  =  .03) 
;r  rod  at  the  i 

Area  of  throttling 


therefore  Wo  =  .03896  sq.in.  is  throttling  area. 
Buffer  rod  at  the  entrance  to  buffer  chamber. 


7X  

-(2.52  -  2.28  )  =  .9048  sq.in, 


Buffer  in  ,394  ina 

.7854(2.36  -  2.28  )  *  2362[1968- (2 .36-2 .28)) 

.3712+. 0276 

a  =  throttling  area  =  .3988 
Buffer  rod  in  .394  +.394  =  .788 


throttling  area  =  .045 


66 


THROTTLIHC  AFEA 

59.99  s  59.93  a 

Clearance  for  the  buffer  rod  rr[(— — )  -  (•  •  • )  i 

25.4  25.4 


.0044 


wx 

Clear  a 

0 

W0  =  .0389 

62.  11 

.0358 

.0389 

62.  "78 

.0315 

.0359 

63.  78 

.  0270 

.031* 

64.78 

.0227 

.0281 

65.78 

.0181 

•* 

.0285 

66.  78 

.0135 

o 

0 

.0179 

67.78 

.0090 

• 

.0134 

68.78 

.0047 

.  0091 

69.78 

0 

.O044 

70.7s 

0 

.0044 

COHSTAMT  ORIFICE  PERIOD. 
ACCELERATING  PERIOD  OF  COUNTER  RECOIL  WITH 


COKSTAKT  ORIFICE. 


pa  -  air  pressure  (Ibs/sq.in) 

Rf  =  floating  piston  friction  (Ibs) 

Aa  =  cross  section  area  of  floating  piston 

(sq.in) 
A  =  effective  area  of  recuperator  piston  (sq. 

in) 
K  =  reciprocal  of  throttling  constant 

w  =  area  ef  constant  orifice  in  valve  of  re- 
o 

cuperator  cylinder  (sq.in) 

2R  =  total  guide  and  recoil  friction  (Ibs) 


3AV  -  SR=irrv  ~  since  the 


Now  [  (p 

throttling 

through 
the  various  orifices  in  the  brake  cylinder  was 


67 


found  to  produce  a  negligible  braking  reaction  as 
compared  with  the  equivalent  throttling  in  the  re- 
cuperator system.    Hence 

dv 

2R=nrV   dJ      Assumin*    a 

uniform  velocity 

attained   at  3   ft/sec.,    we  have   constant   throttling 

area, 

K«A|v2  Rf 

-  -  —  =<Pa  ~  —  )•*  *  ZR 
175  w0  Aa 


175  w« 


K2A3Vg 


Bf 

175[CPa-  —)A-3 


If   K  = 


Ra    =    .253    Pa+.268   ph+    1710 

360+220+1710 
ZR=2300   Ibs. 

1.11      x    6.24      x   3  _    _   18.462 
1761(1420   -   172)    6.25  -   2300] 

=    .1543   x   To 
W      =    .0398   sq.    in. 


Constructively  we   have  WQ  =    .03896   sq.in. 

COMPU1ATION  OF  THE  COUNTER  RECOIL; 

Pf      KV«  dv 

>*  -  -    ;  « 


KV*    -    (.253p.    +   100    +    1710)=   mv   V  ~ 


68 


.352   pa+3430 

-  1810 


KV«  dv 

6.25  pa-.0963  pa-938  -  — .253   pa  -  1810  =  my   V  ~ 

KV2  d? 

5.9  pa 5 275C  =  mv   V  -— 

o 

KV2  i 

5.9  pfl   dx  -~  dx  -  2750  dx  =  -m(V2   Va  ) 

W  SI 

2x5.9Padx   ^2_K7^      2x2750  dx 

284      "  284  w2   dX  "     284      =V2-V2 


.0416   Pa  dx  --  —     V2+V2  dx  -  19.35  dx   =  V«   -  V» 

^"o 

.6064   dx  V2        .6064  dx  V2 

.0416F_   dx  --  "  --  i-  -  19.35  dx=(V«-V  ) 

10«w2 


+  )=  .  ^  _   18>35 

10«W£  10W2 

W§  =    .0389      =    .001513 

V|(l    +    .334   dx")   =  V2(l  -   .334   dx")    +   .00347  padx 
-   1.611   dx" 


69 


C'Fe- 

dx" 

Pa 

.334dx 

.00347PJ 

.00347padx 

l.Sllax 

V 

.25 

.25 

2.  160 

.0835 

7.5 

1.87 

•  402 

1.  165 

.50 

.25 

2.  150 

.0835 

7.45 

1.865 

.  402 

1.58 

1. 

.  50 

2.  145 

.167 

7.43 

3.72 

.605 

2.07 

2. 

1. 

2.  125 

.33* 

7.37 

7.37 

1.  611 

2.54 

4. 

2. 

2.  100 

.67 

7.27 

14.27 

3.  222 

2.  84 

6. 

2. 

2.  060 

.67 

7.15 

14.30 

3.  222 

2.  87 

10. 

4. 

2.OOO 

1.3  3 

6.94 

27.76 

6.45 

2.82 

15. 

5. 

1.950 

1.67 

6.76 

33.80 

8.05 

2.77 

20. 

5. 

1.  90O 

1.67 

6.6 

33. 

8.  05 

2.72 

25. 

5. 

1.  845 

1.67 

6.4 

32. 

8.  05 

2.66 

30. 

5. 

1.  800 

1.67 

6.  25 

31.  25 

8.05 

2.63 

35. 

5. 

1.  730 

1.67 

6. 

30. 

6.05 

2.55 

40. 

5. 

1.  700 

1.67 

5.9 

29.5 

8.O5 

2.53 

44. 

4. 

1.  6*70 

1.35 

5.72 

23.19 

6.45 

2.51 

48. 

4. 

1.  600 

1.33 

5.55 

22.8 

6.45 

2.48 

51. 

3. 

1.  600 

1. 

5.47 

16.  64 

4.83 

2.43 

V* 

2 


(l+.334dxn) 


„. 


Cl-  .334  dx")    .00347padx"-1.611dx" 


C'  Re- 
coil 

dx" 

a 

b 

Pa 

c 

d 

I 

Vs 

V 

51 

3 

200 

0 

1600 

16.  64 

11.  80 

11.  80 

5.90 

2.43 

54 

3 

200 

0 

1580 

16.45 

11.  61 

11.  60 

5.  8C 

2.41 

56 

2 

1.  668 

.332 

1560 

10.  81 

7.59 

9.52 

5.71 

2.39 

58 

2 

1.668 

.332 

1540 

10.67 

7.45 

9.35 

5.  6c 

2.36! 

59 

1 

1.334 

.666 

1530 

5.32 

3.71 

7.44 

5.57 

2.36 

BUFFER  IK  ACTION  -  RETARDED  COUNTER  RECOIL. 


The  dynamic  equation  during  the  buffer  action 
period,  becomes 

Rf    K*A$  v* 

mm 


K«Agv» 


dv 


Kv: 


K  v; 


I  i\v  i 

Simplifying,    we   have    (p     ~  T~)Av~  "Is          It 

fl      w     w 


dv 


70 


Since   ZR=253   p8+100+17lO  we   have 
.352pa+3430  KV»        KtVa 

PaAv Av ~ 253   pa~1810= 

22.85  -«          -* 


dv 

— 

dx 


. 
mr   V  — 


KV*  t  dv 

5-9  pa  -  -  ---  —  =   2750  -  »r  V  — 
w 


o  x 


2*6-9Pa  2    KV2  2   K,v*          2x2750 

-  dx  ---  dx  ---  l  —  dx   -  =  —   dx   =V2-VJ 
284  m   w*  m      w2  284 


•o 

K 


.041padx  -  ID  x  —  V2  +  V2    dx Va+V2    dx   - 

W«          2          *  m»2          "          » 


2x2750 


284 

dx   -   V2    -  Vs 


6064  6064  VV«  +  V2) 

.041   Pa  dx  --     dxV2   --  dx  V»   -  -2—-Z  -  1—  dx 
1C6W|                   10'W2  raw2 

=    IS.  35  dx  -  V2   -  V2 
3 


106 

6064  6064  3804V2 

0416Padx  --  V'dx   --    dx   --  ^-  dx   - 
10«w2      »  1C'*2  10«w2 

3804V2 

...    -1    dx      -   19.35  dx   =   V2-V* 

10eW2  2          1 


+    .0416  p.   dx    -19.35  dx 


71 


00 


ro   <J- 


CT- 


fc.  <> 


X 

-o 


Cu 


8 

O 


CO 
CO 


X 

•a 


M  n 

£> 
O 


CO 

o 
o 


o 

CO 


MX 

* 


X 

pt 


cv«>  « 

I    CfS    .-» 

CM  m 
CM  i-i   o  m  oo  con  r~ 

NO     ON    «n    ON  ^*  O"*     I"4 
NQ     ^"     •»     ^*    •*  K1O    CM 

ON  in  CM   o>-  *  ^ON  CM 

o  QO  x*-  o  ^"*  irtM  o        ON 

to  H  i-t 
o  NO   *»   o 

o   o  m  o  *^mO  in        o          o  in  o 

<N   oo  in  CM  oo\o*  H        oo          10  CM  CM 

W      i-(      ri     rH^QOO               ON                 ON  ON  ON 

in  m  in  inmiom  in        ^          *  «  <* 

•*   ^i  \o  £-°°  OCN  •*         it\         NO  o  * 

O      ON     00    f^^NOlO    *               IO.                CM  CM  rt 

in  •*   ^   •*  "  •#*  *        f         •*  •*  * 

H     i-t     i-l     i-l  *"*  r<»-l    i-l            i-l               ri  r* H 

ON    i~i    r4    X"*^OOO    *H             in              in  tH  rl 

to  in  to  owoo  r«       NO         NO  co  oo 

mx-.e**'tr>CMOio,        o          CM  «*  * 

,,»>•*..          .           .  .  . 

IO  IO 

i-l   ON  ON  to  •n  o<x>  ON        m         m  H  H 

NO   >»  \o   ON^~  oo  or        NO         \o  oo  OD 

<*   CM   CM   in^3  ooo  NO         o          CM  •*  •* 

••••*•••          •            .  •  • 

ri  ri  H  i-**"*  I-KM  i-i        CM         CM  m  in 

rt  oo  r^ 

NO       ON     tO 

^     ^»     Id    in  \o  GOO    IO            T-l               O  ON  ON 

•       •••.•••              «                 *  •  . 

rH    H            CM               in  X1-  X^ 

H  i-l 

x-  i1  to  os-ito**  m       m         in  o  o 

1-4  ri  <N  CMJ^^NO  o        OD         x~  m  m 

•»  x-  x- 

rt fl 

CM  K>  m 

oo  •*  •* 

000 

OOO    Oo  °°   O           O             O  O  O 


CO 
CO 


l-M  CM     1-l|  CM       i-1J   •* 


OC  -H 

-    o 

O   0 


72 


_ 

K   A»V»        1.11    x   6.24 


178*      38 


1.23*36.94  48.07 

'   2646 


175*' 


175x6. 24w« 


ZF 


.352p   +3430         .253pa+1810 

* 


22.66 


6.24 


.0154pa+150.11+.0405pa 
+  280  =    .056pa+440 


.352pa+3430        .253pa+.268   +1710 


22.86 


6.24 


.0154pa+150  +   043   ph 

0483pa+   275 
.0637pa+.043ph+430 


C1  Re- 
coil 

?a 

Ph 

063  7p 

•043Ph 

i 

2150 

830 

138 

36 

604 

2 

2130 

825 

136 

36 

604 

4 

2100 

820 

135 

35 

600 

6 

2060 

815 

132 

35 

^00 

8 

2030 

810 

130 

35 

500 

12 

1980 

800 

12"7 

34 

500 

20 

1900 

V30 

122 

31 

583 

30 

1800 

660 

116 

28 

5*74 

40 

1*700 

600 

109 

26 

565 

50 

1580 

520 

101 

22 

553 

60 

1510 

450 

97 

20 

5<7 

65 

1480 

410 

95 

IV 

542 

68 

1450 

380 

93 

16 

539 

•70 

1440 

3*70 

92 

16 

538 

73 


COMPONENT  FORCES 


C'Fe- 
coil 

pa 

V2 

KV* 

Ph-KV* 

W|10« 

KtV« 

.5 

2150 

2.50 

4.  20 

i7oo 

i. 

2150 

4.  28 

•790 

1360 

2. 

2125 

6.44 

1180 

950 

4. 

2100 

8.  15 

1500 

600 

6. 

2060 

8.  24 

1510 

550 

10. 

2000 

8.01 

1470 

520 

15 

1950 

•7.70 

1420 

530 

to 

1900 

•7.45 

1370 

530 

25 
30 
35 
40 

1845 
1800 

ilio 
1700 

7.  20 

6,00 

**ii 

6.43 

1320 
1270 

\m 

530 
530 

IK 

45 

6.  20 

50 

1580 

6.00 

1040 

480 

55 

5.  74 

60 

1500 

4.53 

830 

67o 

2350 

320 

350 

62 

1490 

3.72 

690 

800 

1560 

420 

380 

64 

1460 

2.  86 

530 

930 

930 

540 

390 

66 

1450 

1.82 

340 

1110 

484 

660 

45C 

68 

1440 

.8"? 

160 

1280 

175 

870 

410 

69 

143O 

.  42 

80 

1350 

72 

1010 

340 

CONSIDERATION  OF  VARIOUS  ELEMENTS  OF 
DESIGN  -  HYDRAULIC  BRAKE  LATOUT. 

In  the  type  discussed,  the  throttling  is  ef- 
fected through  grooves  in  the  counter  recoil  buffer 
rod,  there  being  two  sets  of  grooves,  one  for 
horizontal  elevation  and  long  recoil  and  the  other 
for  max.  elevation  and  short  recoil. 

The  recoil  throttling  is  not  effected 
materially  by  the  filling  in  buffer  since  the 
filling  in  process  is  not  continuous  throughout  the 
recoil.   Hence  the  throttling  grooves  are  computed 
in  the  ordinary  manner.   The  maximum  throttling 
groove  occurs  at  horizontal  recoil  near  the  be- 
ginning of  recoil. 


max 


the  maximum  allowable  hydraulic  pres- 
sure (assume  400C  Ibs/sq.in) 


74 


C  =  the  reciprocal  of  the  contraction  factor 

of  the  throttling  orifice  (assume  C  =  1.38) 

Vr  *  the  max.  restrained  velocity  of  recoil 
(assume  Vr  =  0.9  Vf)(ft/sec) 

wv+4700w 
Vf  =  •          =  max.  free  velocity  of  recoil 

"r          (ft/sec) 
where  w  =  weight  of  projectile  (Ibs) 

w  =  weight  of  powder  charge  (Ibs) 
wr  =  weight  of  recoiling  parts  (Ibs) 

v  =  muzzle  velocity  of  projectile  (ft/sec) 
K  =  total  resistance  to  recoil  at  horizontal 

or  min,  elevation  (resistance  to  recoil 

should  follow  stability  slope  )  (Ibs) 
Ks  =  total  resistance  to  recoil  at  max. 

elevation  (Ibs)  (made  constant 

throughout  the  recoil) 

P  =  total  hydraulic  pull  (Ibs) 
h 

Fv^  =  initial  recuperator  reaction  (Ibs) 
Fyf  =  final  recuperator  reaction  (Ibs) 

Fvf 
®  ~  — —  =  ratio  of  compression  (1.5) 

Fvi 
n  «  coefficient  of  friction  (0.2  to  C.3) 

Then  Fyj  =  1.3Wr(sin  /#*  n  cos  /$)  the  max.  hydraulic 
pull  at  short  recoil,  max.  elevation. 


Phm  =  Ks+wr(sin  0m~O.S  cos  0rnHpvi  (Ibs)  the  irax. 
hydraulic  pull  at  long  recoil,  horizontal  elevation., 

P   =  K  -0.3W  -F.i  (Ibs) 
ho    h 

The  required  effective  area  of  the  recoil  piston 

then  becomes,      p 

ho 
A  =  sq.an. 

Hi  max 
Hence  the  roax.  throttling  orifice,  becoves, 

C  A  Vr 
w      =  — — — — — —    (sq.in) 

h  maX     13.2 


75 


and  the  minimum  throttling  orifice,  becomes, 

C  A  Vr 

W       =  (sq.in) 

n  mm   - 


,,,ax 

The  throttling  areas  decrease  approximately  in 
the  recoil  as  a  parabolic  curve.   For  a  satisfactory 
layout,  for  the  cross  section  of  the  brake  cylinder 
the  inside  cross  section  of  the  hollow  recoil  rod 

should  be  made  approx.  three  times  the  maximum 
throttling  orifice,  that  is  three  times  the  max. 
area  of  the  four  grooves  in  the  counter  recoil  rod. 

Hence  tne  inside  diam.  of  the  hollow  piston 
rod,  becomes, 


/ 


h  max 
— — —   in. 


0.7854 

The  diam.  of  the  spear  buffer  chamber  in  the 
hollow  recoil  piston  rod,  is  obtained  by  a  consideration 
of  the  counter  recoil  stability  and  the  maximum 
allowable  buffer  pressure. 

The  maximum  buffer  reaction  on  counter  recoil 
becomes,         Ws^s 

B'  =  Fvi+Cs— —  -  °'3  Wr 

where  Fvj  =  the  initial  recuperator  reaction  (Ibs) 
Cg  -  constant  of  counter  recoil  stability 

=  0.85 

Ws  =  weight  of  entire  system  (Ibs) 
Wr  =  weight  of  recoiling  parts  (Ibs) 
lg  =  horizontal  distance  from  V&  to  wheel 

and  ground  contact  (in) 
The  max.  allowable  buffer  pressure,  pb'm  =  1500  to 

2500  Ibs/sq.in.    Ha  nee  the  area  of  the  spear  buf- 
fer chamber  in  the  recoil  rod,  becomes, 

B1 
/  * (sq.in) 

Pb'm 

When  counter  recoil  stability  is  not  of  con- 
sideration it  is  customary  to  make  the  buffer  stroke 


76 


about  17.6*  of  the  length  of  horizontal  recoil. 
Assuming  an  entrance  velocity  into  the  buffer  at 
2.5  ft/sec.,  we  have, 

f  nnr  v* 

B'  *  Fvi  *       •  where  v  =  2.5  ft/sec. 
0.175  bh 

=Fvi  0.357  Wr  where  Wr  =  weight  of  the  recoil- 
ing parts,  and  the  area 
of  the  spear  buffer  becomes. 


A  further  consideration  requires  the  diam.  of 
the  buffer  to  be  approximately  a  given  ratio  of 

the  inside  diam.  of  the  recoil  rod,  in  order  that 
a  proper  layout  for  the  filling  in  holes  to  the 
buffer  may  be  made.   Then, 

dib   2  , 

—  =  -  or  dib  =  -  d 

where  0.7864  d*b  =  Ab 

The  proper  thickness  of  the  buffer  rod  is 
based  on  a  consideration  of  both  the  maximum 
longitudinal  tension  during  the  acceleration  or 

powder  period,  and  the  compression  stress  due  to 
an  external  pressure  on  a  hollow  cylinder,  the 

internal  pressure  being  zero.  The  max.  allowable 
fibre  stress  in  the  brake  rod  will  be  taken, 

fffi  =  -  to  -  elastic  limit  (Ibs/sq.in) 
The  outside  diam.  of  the  rod  based  on  max.  tension 
becomes 


0.7854 


where  d_  =  outside  diam.  of  rod 


inside  diam.  of  rod 

weight  of  piston  +  that  portion  of  the 

rod  (estimated  )to  the  farthest  hollow 

section  from  the  piston. 

weight  of  recoiling  parts 


77 


P  =  max.  powder  reaction  on  breech 

b 

P  =  max.  hydraulic  braking. 
h 

WP 

To  allow  for  the  acceleration  load  —  (P  -  B)  we  may 

wr     b 

increase   P     to   1.3   P   ,    then 

~~       ~~~~ 


which  is  sufficiently  close  in  a  preliminary  lay- 

out.  If  we  now  consider  the  compression  due  to 
the  hollow  piston  rod  with  an  external  com- 

pressive  pressure,  we  have,  fm(d0-d.j)=  1* 


di  approx. 

hence  dQ  =  -  where  pbm  =  max.  pres- 
1  -3phm  sure  in  the  brake 

fm     cylinder  (Ibs/sq.in) 

The  diam.  of  the  recoil  brake  cylinder  may  now  be 
calculated.   We  have  0.785  d*  =  A+0.785  d* 

»here 


-0.3  cos 
A  = 


Ph  max. 

KS  =  total  resistance  to  recoil  at  max.  ele- 
vation. 
Ph  max  ~  max-  allowable  pressure  in  recoil 

cylinder  consistent  with  packing 
=  4000  Ibs/sq.in. 

RECUPERATOR  LAYOUT. 


(1 )     Ratio  of  compression  m  : 

The  ratio  of  compression  is 

based  primarily  on  a  consideration  of  proper 
counter  recoil  functioning.   A  high  ratio  of 
compression  decreases  the  size  of  the  recuperator 
and  accelerates  the  counter  recoil  at  max.  elevation, 
while  on  the  other  hand  a  large  value  cf  m  causes 
undue  heating  which  is  injuraous  to  the  packing  in 


78 


the  floating  piston,  and  further  limits  the  horizontal 
stability. 

m  should  be  taken  at  from,  1.5  to  1.7 

(2)     Ratio  of  recuperator  cylinders: 
If  Ay  =  effective  area  of  recuperator  piston  (sq.in) 
.Aa  *  cross  section  of  air  cylinder  (sq.in) 

Paf 
n  =  —  —  *  ratio  of  compression  =  1.5  to  1.7 

Pai 

Aa 
r  =  —  =  ratio  of  recuperator  cylinder 

Av 

1  =  length  of  air  column  in  terns  of  cross  section 
area  of  air  cylinder. 

j  =  -  =  length  of  air  column  in  terms  of  re- 
b 

coil  stroke 

Then,  the  initial  volume, 
i 

k 

V.  =  Av  b  —  -  where  k  =  1.3 


a       t 
and  since  —  =  r  =  *  =  J 


.'        *« 

Let  j   »  1   approx.   hence   r  =     ••  =  —  - 


To  compute  AV  we  have  Fvi  *  1.3Wr 

1.3  Wr 
hence  A^  =       (sin^-m  cos#n)  sq.in. 

pai 

where   Pai   *   1500   Ibs/sq.in.    (approx.) 
Then   Aa   =   Ay    r      sq.in.      A  further   limitation   for   r, 
is   that  y 

r   =  -  where  V,nax  =   max.   vel.    of   recoil 

1  £i 


79 


12  ft/sec.  =  max.  allowable  vol.  of  floating1  piston 

LAYOUT  OF  CALCDLATIOI8  FOR  COUHT1H  RECOIL. 

(l )     During  the  constant  orifice  period, 

Bf    c'A»v«  dv 

Av<  Pa ) ZR*«r  v  — 

Aa    176  •  •  dx 

where  AV  =  effective  area  of  the  recuperator  piston 

(sq.in) 

p  *  air  pressure  (ibs/sq.in) 
Rf  «  floating  piston  friction  (Ibs) 
Aa  =  area  of  cross  section  of  air  cylinder 

(sq.in) 

C  =  reciprocal  of  contraction  factor  of 
constant  orifice  =  — — - • 

w0  =  area  of  constant  orifice  in  recuperator 

system  (sq.in) 
2R  =  total  friction  -  guide  +  packing  (Ibs) 

The  max.  velocity  counter  recoil  is  attained 
shortly  after  the  beginning  of  counter  recoil.  If 
paf  =  final  air  pressure   (Ibs/sq.in) 
paj,  =  initial  air  pressure  (Ibs/sq.in) 
Then  Pa  -  pai+0.9(Paf-pai) 

pai 

=  (9m+l)=0.95  m  paj  approx. 

10 

Paf 
where  m  = =  1.5  to  1.7 


dv 
At  max.  vel.  mrv  —  =0   Assuming  a  max. velocity 

^x       at  horizontal  recoil  = 
3  ft/sec. 

C»A«v8  Rf  C«A»v» 

-  J  —  =  Av^Pa ^  ~  2R  and  ^o  =   - 

176w«  Aa  Rf 

175tAv(pa-  -p)-2 
Aa 


80 


hence  wo  =  .  sq.in, 

13 


f~    Rf 
.2/Av(pa-  —)-] 


The  constant  orifice  should  be  designed  for  counter 
recoil  at  horizontal  or  minimum  elevation. 

As  a  numerical  illustration,  the  value  of  the 
constant  orifice  in  one  model  has  been  worked  out  as 
follows: 

Rf       yua 
(Pa  -  — )AV  -  —  =  2R 

A        W1 
*a      wo 

.352Pa+3430        Kt  AS  v* 
(p )6.24  - =.253  pa+  1810 

22.88  175W» 


6.24Pa-.0961Pa  -  936.68 — =  .253P&+1810 

o 

5.891P,-2746.68  =  15.4267 (P  =2080  Ibs) 

wo 

15.4267 
12253.28  -  2746.68  =  Wo  «          .0016 

WQ  =  .04  sq.in.  from  drawings  =  .03896 

Percentage  accuracy  2.57 

Compute  velocity  of  the  recoiling  parts  when  buffer 

is  entering  V0  =  2.35  ft/sec. 

(2)     Retardation  period  of  counter  re- 
coil. 

In  the  retardation  period,  we 

will  assume  a  constant  unbalanced  force  acting  in 
the  retardation.  Let 

0  =  total  unbalanced  force  in  counter  recoil 

retardation  (Ibs) 
vo  =  entrance  velocity  into  buffer  chamber 

(ft/sec) 
k  =  factor  of  safety  0.95 

C'Agv2 

B1  = 

175  w£b 

lb  =  length  of  buffer  (ft) 

xb  *  displacement  in  buffer  (ft) 

pa  =  pressure  on  oil  side  of  floating  piston  in 


81 


air  cylinder  (Ibs/sq.in) 
p  *  pressure  in  recuperator  (Ibs/sq.in) 
Av  s  effective  area  of  recuperator  piston  (sq. 

in)                 C'*A2  v2 
then  J0=B'+ZR-PvAv  «B'+2R  -(p£  -  -^— )AV 


175w» 


S'+ZR-paAv 


175  wa 


-           m  ' 

"xb      wo 

I^r7!  mr 

Further  /J  =  -rr Clbs)      and  6  xb  =  —r(v0-v|b) 


20  xb   v« 


2 


hence  v*b  =  v*  --  =  rT"^k  1b"xb^  Hence  the 

m  y»      K  JLv 

equation  of 

counter  recoil  during  the  retardation  period,  be- 
comes, 


m  v 

«  Pa  AV-R 

The  only  variable  being  wxb  which  therefore,  becomes 
a  function  of  xb  as  we  should  expect. 

To  solve  for  wxb  we  have  the  following  expres- 
sion in  terms  of  xb  ,  the  displacement  in  the 
counter  recoil  in  the  buffer: 
C'2Af(k  Ib-xb)v0 

A»  (kl-x)   B    v 


C'A^Vp       (  k  lb  -  xb 
hence  Wxb  z  — — ^-^  /  — — — ^— ^— — 

13-2 


175 


82 


•  (sq.in) 

F  »     O  * 

o     If  1 1_ 

Rf 
Obviously  p1  =  pa  ,  as  a  function  of  xb 

and  its  proper  value  must  be  used  for  a  given 
Computation  of  v_: 


Knowing  wo  from  (1),  we  may  readily  calculate 
vo  by  the  logarithmic  method,  as  follows: 

.   C'gA*(x-x) 
" 


200  %wS 
where  A  =  pa  Av  -  2R 

Only   few  intervals  are  needed,  the  mean  air 
pressure  being  taken  for  these  intervals. 


The  following  is  an  illustration  of  actual 
calculations  made  for  the  entrance  into  the  buffer 
on  counter  recoil. 

LAYOUT  CALCUL ATIOHS  OF  BUFFER. 


Vo  •  velocity  of  counter  recoil  when  buffer 
enters 

R  b  c  A  -  ~maV*  =  energy  to  be  dissipated 

during  buffer  action 
where  b  =  length  of  buffer  in  inches  =  12  in. 

R  *  a  factor  of  safety  .95 
-a  V*    -284*2.38 


dv 

jtf  =  -  MV  — 

m     dx   -  ,  .  2x73^5 
xb=;(V«-V«)  »  2.35    284 


65 


5.67  -  .517  xb 


C'recoil 

*b 

V 

V 

60 

1.22 

2.  22 

2.18 

1%8 

62 

2. 

1.9V 

1.93 

1.8 

64 

2. 

1.69 

1.69 

0 

66 

2. 

1.36 

1.31 

3.  V 

68 

2. 

.89 

.93 

5. 

•70.78 

2.68 

0 

0 

0 

FORCES  (Buffer  forcee) 


K'AgV» 


KA»V2 


»(.253Pa+1810)+ 


K'A|V« 


,362Pa+3430 
22.85 


0.253Pa+1810+ 


15x4    " 


175 


-    (6.24Pa-0.0963Pa 
K 


-938 ) 

175  V*' 


V* 


0.263P.+1810+1.0804  (6.3363P.-938 

**  -.  .. 

l.l*x    6.24**v 


0.253Pa+   1810+1.0804  6.3383Pa+938+1132.74  V« 


1.0804 
68.78   *  -   6.0833P.+2740   +  V»    +   1132.74  V* 

W* 
"x 

1.0804  V1 

6.08P.-2570-1132.74V8 


84 


C'R 

V 

V2 

1.08V2 

Pa 

6.08Pa 

1132 

747V| 

*i 

WK 

Actual 
areas  . 

60 

2.  22 

4.53 

4.  90 

152O 

9250 

5550 

1030 

.004*75 

.0438 

.0395 

62 

1.  98 

3.  92 

4.  24 

1500 

9130 

444O 

2020 

.00210 

.  0438 

.0395 

64 

1.  "70 

2.89 

3.  13 

1480 

9000 

32*70 

3000 

.001O2 

.0319 

.0305 

66 

1.35 

1.83 

1.9*7 

14*70 

8940 

2060 

4210 

.  00047 

.  0217 

.0213 

68 

.95 

*.  90 

.9*7 

1450 

8820 

1020 

5150 

•00019 

.0138 

.0130 

FILLIKG  IN  OIL  HOLES  FROM  RECOIL  CYLINDER 


INTO  BUFFER  CHAMBER. 


Area  of  the  filling  in  oil  holes  must  l>e  large 
enough  to  fill  the  hydraulic  cylinder  during  recoil 
period . 

P  =  average  pressure  =  average  hydraulic  pull 

=  720  Ibs/sq.in. 

d  =  weight  of  cubic  ft.  of  hydraulic  oil  »  53 
Ibs/cu.f t. 


max 


maximum  velocity  of  recoil  «  35  ft/sec. 
b  =  length  of  recoil  «  5.9  ft. 
C  -  contraction  factor  =  .85 
A4  *  effective  area  of  recoil  cylinder  26  B<J. 

in.  »  26  sq.in. 
AjJ  =  area  of  the  larger  buffer  chamber  (d*90 

mm)=  .0685  sq.ft. 


P  = 


V  t 


max 


t  = 


MrVmax 
P. A. 


C  VT 


2gP  „  V 


pA 


-'  Abl 


C  W 


2x32.2x720x144    284x36 
53  *  720x26 


.0685  x  59 


(548)2    (.487)2 


.4042 
1 
W2 


85 


.3        .9483       W2 

J_    B  2.1966 
W*        .2846 

W2   «    .1295  W  -    .36 

=    .4042 


4042 

.002523   s.ft, 


355x. 531x85 

W  =    .002523   x   144  *    .3634 

W  from  design  »    .3634 

W  from  draw.   =    .36  div. 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 

Los  Angeles 
This  book  is  DUE  on  the  last  date  stamped  below. 


MAY   7 


1954 


HftR3 


Form  L9-25m-8, '46 (9852) 444 


IT 

UNIVERSI 
U 


'ORNIA 


A     000  768  739     5 


UF 
643 

U58t 

EBgineerhif 
Library 


MJX1UMHI 


SEP      T3 


